Sep 28, 2014 - with supply: 2p- q â 10 = 0 and demand: (p + 10)(q + 30) = 7200. 2.3 15) Break-even ..... -6x - 2y. = -10. - 5x. = - 5 ⨠x = 1. Plugging into equation 1) gives: 1). 1 + 2y. = 5. ⨠y = 2 .... x â4y = 9 â
(â2) â â2x+8y = â18. 2x â8y = 7.
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1st Seminar Business Mathematics Introduction and Overview Basics of Algebra: Exponents, Radicals, Operations with Algebraic expressions Applications Concept of functions Basics of linear functions Additional Material: Harshbarger, Reynolds: Chapter 0. Odd numbered questions from exercises of sections 0.3, 0.4, 0.5, 0.6, and 0.7 (Solutions at pp A1 – A4) Harshbarger, Reynolds: Chapter 1, Sections 1.2, 1.3. Odd numbered questions from exercises of sections 1.2, and 1.3 (Solutions at pp A4 – A9)
Overview •
Review: Basic algebra
•
Concept of functions, Basics of linear functions
•
Applications of linear functions: Breakeven analysis Market equilibrium analysis o Homework: Linear functions and applications
•
Basics and applications of quadratic functions,
•
Exponential and logarithmic functions o Homework: Quadratic, exponential, and logarithmic functions
Self-Evaluation exam 1 •
Systems of linear equations
•
Basic algebra
Self-Evaluation exam 2 •
Future value of annuities, Present value of annuities
Homework: Applications of finance mathematics
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Exponents and Radicals (pp 15, 20) •
Basically: rules of exponents are valid for radicals, as they can be represented as exponents. NB conditions (p 20).
n a n an n−m n n n a a • a = a ; m = a ; a • b = (ab) ; n = ; (a m ) n = a m•n a b b n n n bc = b ⋅ c if b, c ∈ℜ n
€
n
m
n +m
b nb = c nc
if b, c ∈ℜ and c ≠ 0
1 1 − b m n = b n b
m
m
if b ∈ℜ and b ≠ 0 1 1 + n
b ⋅ n b = bm
b = m⋅n b • Exponential Equations: Solve by taking the logarithm at both sides of the equation. 5x = 1024 ⇒ log(5x ) = log(1024) log(1024) x • log(5) = log(1024) ⇒ x = ≅ 4.31 log(5) m n
5x = 527 ⇒ x = 27
Algebraic Expressions (pp 27) • Add, subtract, and multiply polynomials. (x + 3) + (2xy − 3x + 7) = −2x + 2xy +10 (2x 2 − 7x + 20) − (4x 2 −12x +10) = −2x 2 + 5x +10 (x 2 − 7x)(3x 2 − 2x) = 3x 4 − 21x 3 − 2x 3 +14x 2 = 3x 4 − 23x 3 +14x 2 (x ± 4)2 = x 2 ± 8x +16 (x + 4)(x − 4) = x 2 −16 • Division of polynomials to simplify an expression (p 32) (4x 3
−13x − 22) : (x − 3) = 4x 2 +12x + 23+
47 x −3
x≠0
−4x 3 +12x 2 +12x 2 −13x − 22
•
−12x 2 + 36x 23x − 22 − 23x + 69 47 Factoring (pp 37)
5x − 5y − bx + by = 5(x − y) − b(x − y) = (5 − b)(x − y) x 2 − 7x + 6 = (x + a)(x + b) = x 2 + (a + b)x + ab ⇒ a + b = −7 and ab = 6 ⇒ a = −1 and b = −6 alternatively: find solution through guessing
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Applications in class (p 19, pp 26, p 35, pp 41, p 49, pp 53) 69) Health and care expenditures (0.3 p 20) The national health care expenditure H (in billions of $) can be modelled by the formula H = 30.58(1.102)t , t: number of years past 1960. a) What t-value corresponds to 1970? b) Approximate the national health care expenditure in 1970. c) Approximate the national health care expenditure in 2005. d) Estimate the national health care expenditure in 2015.
68) Life span (0.4 p 26) Life expectancy in the USA can be approximated with the equation L = 29x 0.21 , x: number of years that the birth year is past 1900. a) Express this equation with radical notation. b) Estimate life expectancy for a person born in 2015.
71) Half-life (0.4 p 27) The half-life of strontium-90 is 25 years. Find the amount of strontium-90 remaining after 10 years if q0 = 98 kg.
75) Company growth (0.4 p 27) The growth of a company can be described by the equation N = 500 ( 0.02 ) years the company has been in existence, N: number of employees. a) What is the number of employees when t = 0? b) What is the number of employees when t = 5?
0.7t
, t: number of
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67) Revenue (0.5 p 35) A company sells its product for $55 per unit. Write an expression for the amount of money received (revenue) from the sale of x units of the product.
68) Profit (0.5 p 35) Suppose a company’s revenue R (in $) from the sale of x units of its product is given by R = 215x Suppose further that the total costs C (in $) of producing x units is given by C(x) = 65x + 15,000 a) If profit is revenue minus cost, find an expression for the profit from the production and sale of x units. b) Find the profit received if 1000 units are sold.
71) Investments (0.5 p 35) Suppose that you have $4000 to invest, and you invest x dollars at 10% and the remainder at 8%. Write expressions in x that represent a) the amount invested at 8% b) the interest earned on the x $ at 10% c) the interest earned on the money invested at 8% d) the total interest earned.
63) Consumer expenditure (0.6 pp 41) The consumer expenditure for a commodity is the product of its market price, p, and the number of units demanded. Suppose that for a certain commodity, the consumer expenditure is given by 10,000 p −100 p 2 . a) Factor this in order to find an expression for the number of units demanded. b) Use a) to find the number of units demanded when the market price is $38.
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65) Revenue (0.6 p 42) Revenue R from the sale of x units of a product is found by multiplying the price by the number of items sold. a) Factor the right side of R = 300x − x 2 b) What is the expression for the price of the item?
55) Time study (0.7 p 49) Workers A, B, and C can complete a job in a, b, and c hour, respectively. Working together, 1 1 1 they can complete + + of the job in 1 hour. Add these fractions over a common a b c denominator to obtain an expression for what they can do in 1 hour, working together.
57) Average cost (0.7 p 49) Average cost = total cost / x 4000 + 55 + 0.1x x a) Express the average-cost formula as a single fraction. b) Write an expression that gives the company’s total costs.
A company’s average costs are given by Average cost =
58) Average cost (0.7 p 49) Same as No 57) with Average cost =
40,500 +190 + 0.2x x
60) Annuity (0.7 p 49) (1+ i)n+1 −1 −1 . i Write the expression over a common denominator and factor the numerator to simplify.
The formula for the future value of an annuity due involves the expression
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Solutions 0.3 69) a) 10,
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b) $80.771 billion,
0.4 68) a) L = 29100 x 21 ,
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c) $2418.9 billion,
d) $6388.9 billion
b) L = 29 ⋅1150.21 = 78.55 years (when was the model created?)
−t −10 0.4 71) q(t) = q0 2 k = 98 2 25 = 74.3 kg ,
0.4 75) a) N = 500 ( 0.02 )
0.70
= 10 , b) N = 500 ( 0.02 )
0.75
= 259.01 ≈ 259
0.5 67) 55x 0.5 68) a) P(x) = 215x – (65x + 15,000) = 150x – 15,000 b) P(1000) = 150,000-15,000 = $135,000 0.5 71) a) 4000 – x, b) 0.10x,
c) 0.08(4000 – x),
d) 0.10x + 0.08(4000 – x)
0.6 63) a) p(10,000 −100 p) ⇒ x = 10000 −100 p , b) x = 10000-3800 = 6200 0.6 66) a) R = x(300 − x) , 0.7 55)
b) 300 – x
1 1 1 ab + ac + bc + + = a b c abc
0.7 57) a) Average cost =
4000 + 55x + 0.1x 2 , x
0.7 57) a) Average cost =
40,500 +190x + 0.2x 2 , b) Total cost = 40,500 +190x + 0.2x 2 x
b) Total cost = 4000 + 55x + 0.1x 2
n (1+ i)n+1 −1− i (1+ i)n+1 − (1+ i) (1+ i)(1+ i) −1 0.7 60) = = i i i
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Concept of functions (pp 71) Relation: set of ordered pairs and rule how to find these pairs Function: relation between two or more sets. For each input element there is exactly one output element. Input: the independent variable. Output: the dependent variable. It is not a function, if more than one output element corresponds to one input element. Notation: y = f(x). f(x) corresponds to the rule how x is operated to receive a value of y and f(x) is what we see when we graph the function. ! Graphing points into a coordinate system ! Graphing a function using a value – table Example: Graph the function f(x) = y → f(x) = 2x2-3x+1 (compare with Example 2, p. 74) Value Table: x 0 y 1
1 0
2 3
3 10
4 21
5 36
Graph:
Verify the values of y in the value table and the graph. Domain: values of x, Range: values of y. The domains are restricted whenever values as 0 in the denominator or even roots of negative numbers result in the range (Example 7 p. 77). Operations with functions: sum, difference, product, quotient (denominator ≠ 0) are possible. Example: f(x) = 2x + 3, g(x) = 2x2 – 4 (compare with Example 8 p. 78) Compute: a) ( f + g)(x);
b) ( f − g)(x);
a) ( f + g)(x) = 2x + 3+ 2x 2 − 4 = 2x 2 + 2x −1;
c) ( f ⋅ g)(x);
f d) (x) g
b) ( f − g)(x) = 2x + 3− (2x 2 − 4) = −2x 2 + 2x + 7 f 2x + 3 c) ( f ⋅ g)(x) = (2x + 3)⋅ (2x 2 − 4) = 4x 3 + 6x 2 − 8x −12; d) (x) = 2 , if x 2 ≠ 2 resp. x ≠ ± 2 2x − 4 g
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Composite functions
g of f (g f ) and f of g ( f g) → (g f )(x) = g( f (x)) and ( f g)(x) = f (g(x)) Examples: 1) f(x) = 2x3 + 1, g(x) = x2 (compare with Example 9 p. 79) Find: a) (g f )(x); b) ( f g)(x)
a) (g f )(x) = g( f (x)) = g(2x 3 +1) = (2x 3 +1)2 = 4x 6 + 4x 3 +1 b) ( f g)(x) = f (g(x)) = f (x 2 ) = 2(x 2 )3 +1 = 2x 6 +1
2) f(x) = 1 – 2x, g(x) = 3x2 (compare with Checkpoint p. 79) Find: a) (g f )(x); b) ( f g)(x)
a) (g f )(x) = g( f (x)) = g(1− 2x) = 3(1− 2x)2 = 12x 2 −12x + 3 b) ( f g)(x) = f (g(x)) = f (x 2 ) = 1− 2(3x 2 ) = −6x 2 +1
Basics of linear functions (pp 85)
General form of a linear function: f ( x) = mx + b . A linear function has a slope (m), an intercept with the y-Axis (y-Axis-Intercept = b), and with the x-Axis (null). Both intercepts b lead to a specific point of the linear function, namely: x = 0 ⇒ (0;b); y = 0 ⇒ − ;0 . These m two points allow drawing the graph of the linear function. 5 Example: y = 2x + 5 x = 0 ⇒ (0;5); y = 0 ⇒ − ;0 . Now draw both points and connect them. 2
y2 − y1 Δy . = x2 − x1 Δx Directions of slopes: positive (m > 0), negative (m < 0), horizontal (m =0), vertical (m not 1 defined / included). Two lines l1 and l2 are perpendicular, if m1 = − . m2 The slope m indicates the rate of change (also for non-linear functions). m =
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Example: Give the linear function, which contains the point (2; - 3) and has a slope of -0.5.
m = −0.5 ⇒ y = −0.5x + b; P = (2;−3) ⇒ −3 = −0.5⋅ 2 + b ⇒ b = −2 ⇒ y = −0.5x − 2 (p 89) Example 5 (p 90) Average price p of digital television can be expressed as linear function of the number of sets sold N (in thousands). If N increases by 1000, p drops by $10.40, and when 6485 are sold, the average price set is $504.39. Give the linear function from the information above.
Δp −10.40 = = −0.0104;Point: (6485;504.39) ⇒ 504.39 = −0.0104 ⋅ 6485 + b ΔN 1000 ⇒ b = 571.834 ⇒ p(N ) = −0.0104N + 571.834 Example: Solve for the slope-intercept form (slope and intercept can be directly seen in the function). 5x – 20y = 100 (p 92). p = mN + b ⇒ m =
20 y = −5 x + 100 ⇒ y = 0.25 x − 5 General form: ax + by + c = 0; Point-slope form: y − y1 = m(x − x1 ); Slope-intercept form: y = mx + b Vertical line: x = a; Horizontal line: y = b
Applications in class (pp 80, pp 93) 1.2 55) Profit The profit from the production and sale of x units of a product is given by: x2 P(x) = 180x − − 200 . Suppose that for a certain month the number of units produced on 100 day t of this month is: x = q(t) = 1000 +10t a) Find (P q)(t) to express the profit as a function of the day of the month. b) Find the number of units produced, and the profit, on the fifteenth day of the month.
1.2 59) Fencing a lot A farmer fences the perimeter of a rectangular lot with an area of 1600 square feet. The lot is x feet long. Express the amount L of fence needed as a function of x.
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1.3 49) Depreciation A $360,000 building is depreciated by its owner. The value y of the building after x months of use is y = 360,000 – 1500x. a) Graph this function for x ≥ 0. b) How long will it take until the building is completely depreciated (value = zero)? c) The point (60; 270,000) lies on the graph. Explain what it means.
1.3 57) Electric costs An electric utility company determines the monthly bill for a residential customer by adding an energy charge of 8.38 cents per kilowatt-hour (x) to its base charge of $16.37 per month. Write an equation for the monthly charge y in terms of x.
1.3 61) Consumer price index (CPI) The CPI for urban consumers for the years 1985 – 2005 can be accurately approximated by a linear model determined by the line connecting the points (1985; 113.2) and (2005; 324.9) with x: year and y: price consumers pay in year x for goods that cost $100 in 1982 (base year). a) Write the equation of the line connecting these two points to find a linear model for these data. b) Forecast the CPI for the year 2020!
Solutions 1.2 55) a) ( P ! q)(t ) = 180(1000 + 10t ) −
(1000 + 10t ) 2 − 200 ; 100
b) x = 1150, P = $193,575
2 ⋅1600 ⎛L ⎞ 1.2 59) A = 1600 = x(b − x); L = 2b + 2 x ⇒ 1600 = x⎜ + x ⎟ ⇒ L = + 2x x ⎝2 ⎠ 1.3 49) b) 240 months; c) After 60 months, the value of the building is $270,000. 1.3 57) y = 0.0838x + 16.37 1.3 61)
a) P1 (1985;113.2) ⇒ 113.2 = 1985m + b; P2 (2005;324.9) ⇒ 324.9 = 2005m + b 20m = 211.7 ⇒ m = 10.858; b = 113.2 −1985⋅10.858 = b = −20,898.025 y = 10.585x − 20,898.025 b) y(2020) = 10.585⋅ 2020 − 20,898.025 = 483.675
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2nd Seminar Business Mathematics Applications of linear functions: Breakeven analysis Market equilibrium analysis Basics and applications of quadratic functions Additional Material: Harshbarger, Reynolds: Chapter 1, Section 1.6. Odd numbered questions from exercises of section 1.6 (Solutions at pp A9 – A12) Harshbarger, Reynolds: Chapter 2, Sections 2.2, 2.3. Odd numbered questions from exercises of sections 2.2, 2.3 (Solutions at pp A12 – A15)
Breakeven Analysis (pp 119) C(x) = Total cost (fixed cost + variable cost) = cfix + cvar R(x) = Total revenue from sale of x units = (price per unit) • (number of units) P(x) = Profit from sale of x units P(x) = R(x) – C(x) Break-even point: cost = revenue:
R(x) = C(x)
Marginal profit: MP = profit of one additional unit (slope of profit function) Marginal cost: MC = cost to produce one additional unit (slope of cost function) Marginal revenue: MR = revenue of selling one additional unit (slope of revenue function) Example: A manufacturer sells IPads for $400 each. The quarterly production and sale costs (fixed costs) for the IPads are $500,000. Additionally each IPad costs $30 to produce (compare with Examples 1, 2, 3, and 4, pp 119). a) Derive the cost, revenue, and profit functions for the IPads. b) Derive marginal cost (MC), and marginal revenue (MR) c) Derive marginal profit (MP) between the 200,001th and the 200,000th unit. What is the total profit at 200,000 units? d) How many IPads must the manufacturer produce each month to break even? e) What are the total costs at the break-even point? f) Graph the cost, revenue and profit functions, show the break-even point in the graph.
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a) C(x) = 500,000 + 30x;
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R(x) = 400x; P(x) = 400x – (500,000 + 30x) = 370x – 500,000
b) MC = 30$;
MR = 400$
c) 370$ as MP is linear in this case;
P(x) = 370 • 200,000 – 500,000 = 73.5M $
d) C(x) = 166,667 + 30x = R(x) = 400x ⇒ x = 450.45 e) C(451) = R(451) = 400 • 451 = 180,400$ f)
Market Equilibrium Analysis (pp 122) At market equilibrium demand = supply (market cleared). Law of demand: as demand increases the price increases as well. Law of supply: as supply increases, the price decreases as well. For convenience, economists draw the quantity demanded or supplied (q) as the independent variable and the price (p) as the dependent. At market equilibrium, the equilibrium quantity and the equilibrium price apply. Demand and supply curves intersect. To find any equilibrium (BEP or market equilibrium), a system of equations has to be solved. Example: A small towns’ consumers buy 100 IPads per month, if they cost 400$ and 200, if they cost 300$. The manufacturer supplies 50 IPads, if he can sell them for 300$ and 100 IPads, if he can sell them for 400$. Linear demand and supply functions are assumed (compare with Examples 5, 6, and 7, pp 123). a) Find the market equilibrium b) Find the market equilibrium, if the manufacturer is taxed with 10$ per unit sold. c) Graph the demand and supply functions before and after the tax has been issued.
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400 ! 300 = !1 " p ! 300 = !1(q ! 200) " p(q) = !q + 500 100 ! 200 300 ! 400 Supply: m= = 2 " p ! 300 = 2(q ! 50) " p(q) = 2q + 200 50 !100 Equilibrium: p(q) = !q + 500 = p(q) = 2q + 200 " 3q = 300 " q = 100; p = 400
a)
Demand:
m=
b)
Demand: Supply: Equilibrium:
p(q) = !q + 500 p(q)Tax = 2q + 200 +10 = 2q + 210 p(q) = !q + 500 = p(q) = 2q + 210 " 3q = 290 " q = 96.7; p = 403.3
c)
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Applications in class (pp 126) 1.6 9) Cost, Revenue, Profit Let C(x) = 5x + 250, R(x) =27x. a) Write the profit function P(x). b) What is the slope of the profit function? c) What is the marginal profit? d) Interpret the marginal profit?
1.6 11) Marginal Profit A company charting its profits notices that the relationship between the number of units sold, x, and the profit, P, is linear. If 200 units sold results in $3100 profit and 250 units sold results in $6000 profit, write the profit function for this company. Find the marginal profit.
1.6 13) Cost, Revenue, Profit Extreme Protection, Inc. manufactures helmets for skiing and snow boarding. The fixed costs for one model of helmet are $6600 per month. Materials and labour for each helmet of this model are $35, and the company sells this helmet to dealers for $60 each. a) For this helmet, write the function for monthly total costs. b) Write the function for total revenue. c) Write the function for profit. d) Find C(200), R(200), and P(200) and interpret each answer. e) Find C(300), R(300), and P(300) and interpret each answer. f) Find the marginal profit and write a sentence that explains its meaning.
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1.6 15) Find solutions out of the graph (p 127) a) Label each function correctly. b) Determine the fixed costs. c) Locate the break-even point and determine the number of units sold to break even. d) Estimate marginal cost and marginal revenue.
1.6 19, 21) Break-even for Belts A manufacturer sells belts for $12 per unit. The fixed costs are $1600 per month, and the variable costs are $8 per unit. 19 a) Write the equations of the revenue and cost functions. 19 b) Find the break-even point. 21 a) Write the profit function for problem 19. 21 b) Set profit equal to zero and solve for x. Compare this x-value with the breakeven point from problem 19.
1.6 33) Surplus or shortfall If the demand for a pair of shoes is given by 2p + 5q = 200 and the supply function for it is p – 2q = 10, compare the quantity demanded and the quantity supplied when the price is $60. Will there be a surplus or shortfall at this price?
1.6 47) Find the market equilibrium Find the equilibrium point for the following supply and demand functions. Demand: p(q) = !4q + 220 Supply: p(q) = 15q + 30
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1.6 49) Find the market equilibrium A group of retailers will buy 80 televisions from a wholesaler if the price is $350 and 120 if the price is $300 for each. The wholesaler is willing to supply 60 if the price is $280 and 140 if the price is $370. Assuming the resulting supply and demand functions are linear, find the equilibrium point for the market.
Solutions 1.6 9) a) P(x) = 22x – 250; b) 22; c) MP = 22; d) Each unit sold adds $22 to profits at all levels of production, so produce and sell as much as possible. 1.6 11) P = 58x – 8500, MP = 58 1.6 13) a) C(x) = 35x + 6600; b) R(x) = 60x; c) P(x) = 25x – 6600; d) C(200) = $13,600, cost of producing 200 helmets; R(200) = $12,000, revenue from sale of 200 helmets; P(200) = -$1600, loss from producing and selling 200 helmets. e) C(300) = $17,100, cost of producing 300 helmets; R(300) = $18,000, revenue from sale of 300 helmets; P(300) = $900, profit from producing and selling 300 helmets. 1.6 15) a) Revenue passes through the origin; b) $2000; c) 400 units; d) MC = 2.5; MR = 7.5. 1.6 19) a) R(x) = 12x; C(x) = 8x + 1600;
b) 400 units.
1.6 21) a) P(x) = 4x – 1600; b) x = 400 units to break even. 1.6 33) 16 demanded, 25 supplied; surplus. 1.6 47) q = 10; p = $180. 1.6 49) q = 100; p = $325.
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Quadratic Functions Parabolas (pp 151) General form: f (x) = ax 2 + bx + c For f (x) = ax 2 vertex a is turning point, minimal for a > 0, hence opens upward and maximal for a < 0, hence opens downward.
# b # b && Vertex point for f (x) = ax 2 + bx + c ! ... ! % " ; f % " (( setting y = 0 gives two null $ 2a $ 2a '' (intercepts). Because of symmetry the vertex has to be in the middle of these two points. Example: Maximise profit a) Maximise the profit and find the ideal number of units produced for the following profit function (compare with Examples 1, 2, and 3, pp 153).
P(x) = !0.1x 2 + 300x !1200 as a < 0 vertex is maximum point b) Find the intercepts with the x-axis c) Graph the parabola
a) x = !
300 2 = 1500 # P(1500) = !0.1" (1500 ) + 300 "1500 !1200 = 223,800 2 " 0.1
Vertex: (1500;223,800 ) 1500 units provide $223,800 profit. 2 b) P(x) = !0.1x 2 + 300x !1200 = 0 " x1,2 = !300 ± 300 ! 4 # (!0.1)# (!1200) " x1 = 4.00; x2 = 2995.99
2 # (!0.1)
c)
Parabolas do not have “slopes” as linear functions, instead they have “average rate of change” as all non-linear functions. Similar to linear functions, the average rate of change can be found between two values of x, for example x1 = a, and x2 = b applying the “slope” formula: f (b) ! f (a) . The secant connecting the two points (a; f(a)), and (b; b!a f(b)) gets closer and closer to the actual function the nearer a is at b. Average rate of change =
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Example: continued from above … d) Compute the secant for the two x-values: x1 = 0, and x2 = 1000. e) Draw the secant into the graph. f) Compute the average rate of change between these two points. g) Is this average rate of change higher or lower than the rate of change between the two xvalues: x1 = 100, and x2 = 500?
d) P(0) = !0.1" 0 2 + 300 " 0 !1200 = !1200; P(1000) = !0.1"1000 2 + 300 "1000 !1200 = 198,800 e)
f) Average rate of change =
198,800 ! (!1200) = 200 1000 ! 0
123,800 ! 27,800 = 240 it is lower compared to this rate of 500 !100 change. The parabola is steeper at low x-values and becomes flatter towards the vertex.
g) Average rate of change =
Business Applications of Quadratic Functions (pp 161) Quadratic or mixed supply and demand curves at 1st quadrant. Solving these equations leads to a system of quadratic equations. Similarly quadratic cost curves are used for break – even analysis. For monopoly markets: Revenue depends on price and demand ⇒ R(x) = p • x. Examples: The supply function for IPhones is given by p = q2 + 100 and the demand function is p = - 20q + 2500 (compare with Examples 1, 2, 3 and 4, pp 162). a) Find the market equilibrium b) Find the break-even point, if the cost function for the IPhones is given by: C(x) = 360 + 40x + 0.1x2 and the revenue function is R(x) = 60x. c) Find the break-even point in a monopoly market for C(x) = 3600 + 100x + 2x2 and demand for the same period (weekly) is p = 500 - 2x.
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a)
q 2 +100 = 20q + 2500 ! q 2 " 20q " 2400 = 0 ! q = 40 (q = -60 is no solution) p(40) = 20 ! 40 + 2500 = 1700 " vertex: (40; 1700)
b)
360 + 40x + 0.1x 2 = 60x ! 0.1x 2 " 20x + 360 = 0 ! x1 = 20; x2 = 180 C(20) = R(20) = 1200; C(180) = R(180) = 10,800 ! vertices: (20; 1200); (180; 10,800)
c)
R(x) = p ! x = (500 " 2x)x = 500x " 2x 2 R(x) = C(x) ! 500x " 2x 2 = 3600 +100x + 2x 2 ! 4x 2 " 400x + 3600 = 0 x1 = 90; x2 = 10 ! R(90) = 28,800; R(10) = 4800 vertices at (10; 4800);(90;28,800)
Applications in class (pp 158, pp 167) 2.2 31) Profit The daily profit from the sale of a product is given by P(x) = 16x ! 0.1x 2 !100 $. a) What level of production maximises profit? b) What is the maximum possible profit?
2.2 37) Photosynthesis The rate of photosynthesis R for a certain plant depends on the intensity of light x, in lumens, according to R = 270x ! 90x 2 . Sketch the graph of this function, determine the intensity that gives the maximum rate, and determine the maximum rate.
2.2 43) Apartment Rental The owner of an apartment building can rent all 50 apartments if she charges $600 per month, but she rents one fewer apartment for each $20 increase in monthly rent. a) Construct a table that gives the revenue generated if she charges $600, $620, $640. b) Does her revenue from the rental of the apartments increase or decrease as she increases the rent from $600 to $640? c) Write an equation that gives the revenue from rental of the apartments if she makes x increases of $20 per rent. d) Find the rent she should charge to maximise her revenue.
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2.3 7) Equilibrium Find the equilibrium quantity and price for a commodity with demand: p2 + 4q = 1600 and supply: 300 – p2 + 2q = 0
2.3 11) Equilibrium Find the equilibrium quantity and price for a commodity with supply: 2p - q – 10 = 0 and demand: (p + 10)(q + 30) = 7200.
2.3 15) Break-even points The total costs for a company are given by C(x) = 2000 + 40x + x2 and the total revenues are R(x) = 130x. Find the break-even points.
2.3 23) Monopoly profit In a monopoly market, the demand for a product is p = 175 – 0.5x, the revenue function is R = px (x = number of units sold). What price maximises revenue?
2.3 31) Find the market equilibrium fixed costs of a company are: $28,000; variable costs are 0.4x + 222 dollars per unit; selling price is 1250 – 0.6x dollars per unit. a) Find the break-even points. b) Find the maximum revenue. c) Derive the profit function from cost and revenue functions and find maximum profit. d) What price will maximise the profit?
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Solutions 2.2 31) a) 80 units;
b) $540
2.2 37) Intensity = 1.5 lumens; R = 270 !1.5 " 90 ! (1.5)2 = 202.5 2.2 43) a) Table below;
b) Increase;
Rent 600 620 640
Revenue $30,000 $30,380 $30,720
Number Rented 50 49 48
c) R = (50 – x)(600 + 20x); d) $800
2.3 7) q = 216.67; p = $27.08 2.3 11) q = 90; p = $50 2.3 15) x1 = 40 units; R(40) = 130•40 = $5200; x2 = 50 units; R(50) = 130•50 = $6500 2.3 23) $87.50, R(87.50) = $15,312.5 2.3 31) a) x1 = 28 units; x2 = 1000 units; b) $651,041.67 ($650,929.6); 2 c) P(x) = – x + 1028x – 28,000; maximum profit is at $236,196; d) $941.60
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3rd Seminar Business Mathematics Exponential and logarithmic functions Review and Practice: Functions Additional Material: Harshbarger, Reynolds: Chapter 5, Sections 5.1, 5.2, and 5.3. Odd numbered questions from exercises of sections 5.1, 5.2, and 5.3 (Solutions at pp A29 – A33) Chapters 1, 2, and 5: Solve Chapter Tests of Chapters 1 (p 135 – 136), 2 (p 198 – 200), and 5 (p 398 – 399). Do not solve “modelling” questions (for example No 16 p 199, No 30 p 399).
Exponential functions (pp 357) General form: f (x) = a x with a 0 The following functions are also exponential functions: f (x) = ca x ; f (x) = ca x ; f (x) = ca bx Exponential growth function: f (x) = Ca x , y-intercept at (0, C), domain: x
, range: y > 0.
If 0 < a < 1, exponential decay applies (radioactive decay, decay of an advertising campaign, inflation on purchasing power effect). Example 1: Future value of an investment If $5,000 is invested at 4%, compounded quarterly, then the future value S after x years is given by: S(x) = Ca 4 x (compare with Example 2, p 359). 1a) Find the actual growth function. 1b) Find the future value of the investment after 10 years.
1 For continuously compounded interest e is used as basis as e = lim 1+ n n infinitively small time step and infinitively many of these time steps.
n
. For n being an
Example 2: Purchasing power P of a fixed income I of $50,000 per year after t years of 5% inflation can be modelled as: P(t) = I e 0.05t (compare with Example 5, p 362). 2a) Find the actual purchasing power function. 2b) Find the purchasing power after 15 years.
Example 3: From the following exponential function f (x) = ce x , c is given as: c = 5. What happens, if ………. ? (compare with Example 6, p 363). 3a) …….. c’ = - c 3b) …….. c’ > c
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3c) …….. c’ < c Solutions from in class examples 1a) S(x) = 5,000 (1.04)
4x
1b) S(10) = 5,000 (1.04) 2a) P(t) = 50,000 e
4 10
= 24,005.103
0.05t
2b) P(15) = 50,000 e
0.0515
= 50,000 0.472 = $23,618.33
3a) c’ = - c: the function mirrors at the x-axis 3b) c’ > c: the function is c’-c entities higher than the original one. 3c) c’ < c: the function is c’-c entities lower than the original one. Logarithmic functions (pp 369) Logarithmic functions solve problems where the unknown is the exponent. for y = log a (x) with a > 0 and a 1 (logarithmic form) x = a y (exponential form) (logarithmic and exponential functions are inverse to each other, as one can be derived from the other by exchanging x and y and solve for y afterwards. Inverse functions have the same shape mirrored at the function y = x. Change of base of logarithmic functions can be used to simplify / solve. For example: log(x) ln(x) log b (x) = = log(b) ln(b) Example 1: If $P is invested for t years at interest rate r, compounded continuously, then the future value of the investment is given by S(t)=Pert. The doubling time for this investment can be found by solving for t, as S = 2P 2 =ert (compare with Example 4, p 371). 1a) Solve for t. 1b) How long will it take to double if r = 10%?
Example 2: Suppose that after a company introduces a new product, the number of months m 40 before its market share is s % can be modelled by: m = 20 ln . When will that product 40 s have a market share of 35%? (compare with Example 5, p 372).
Example 3: The Richter scale is used to measure the intensity of an earthquake. The magnitude I on the Richter scale of an earthquake of intensity I is given by: R = log with I0 = minimal I0 intensity used for comparison (compare with Example 15, p 378). 3a) Find R if I is 3,160,000 times as great as I0. 3b) The 1964 Alaskan earthquake measured 8.5 on the Richter scale. Find the intensity.
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1 unit of the Richter scale means that the intensity is 10 times greater. Solutions from in class examples ln(2) 6.93 years 0.1 40 40 = 20 ln = 41.6 months 2. m = 20 ln 5 40 35
1a) t =
ln(2) ; r
1b) t(0.1) =
3a) R = log
I I = 3,160,000 for I0 I0
3a) R = log
I for R = 8.5 I0
R = log (3,160,000) 6.5 ;
8.5 = log
I I0
I = 108.5 I0
316,000,000,000
Applications of Exponential and Logarithmic functions (pp 382) Solve exponential equations by using properties of logarithms. Isolate the exponent, take the logarithm on both sides, solve for the variable. Example 1: Company XY finds that its daily sales fall after the end of an advertising campaign. The decline is such that the number of sales is: S(t) = 2000 ( 2 0.1t ) with t = number of days after the campaign ends (compare with Examples 1, 2, 3, 4, and 5 pp 383). 1a) How many sales will be made 10 days after the end of the campaign? 1b) Company XY starts a new campaign when the sales drop below 350. When will that be? 1c) Company XY wants to grow within the next 5 years from 2,000 employees to 2,500. For simplicity the following growth formula applies to their growth: E(t) = E0 e bt how many employees will be occupied at Company XY after 10 years (compare with p 386)? 1d) Their demand function for q in 1,000 units of their main product is: p(q) = 30 3
q 2
.
At what price per unit will the demand = 4,000 units? How many units are demanded if the price is $17.32? 1e) Compute their total revenue when q units are sold based on the following demand function p(q) = 100 e $1,000).
q 10
. Compute their revenue, if they sell 30 units (q in number of units, p in
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Solutions from in class example 1a) S(10) = 2000 ( 2
0.110
1b) S(t) = 350 = 2000 ( 2
) = 1000
0.1 t
35 =2 200
)
0.1 t
log(0.175) = 0.1t log(2)
0.1t =
log(0.175) log(2)
t = 25.15
They should start a new campaign before the 26th day. ln(1.25) 2500 = 2000 e b 5 1.25 = e b 5 = b = 0.045 1c) 5 E(10) = 2000 e 0.04510 = 3136.6 3137 employees 1d) p(q) = 30 3
4 2
= $3.33
p(q) = 17.32 = 30 3
q 2
0.5773 = 3
1e) R(q) = q p(q) = 100 q e
q 10
q 2
ln(0.5773) =
R(30) = 100 30 e
q ln(3) 2 30 10
q=
2 ln(0.5773) 1,000 ln(3)
= $149.40
Applications in class (pp 366, pp 167) 5.1 32) Compound Interest If $1000 is invested for t years at 10%, compounded continuously, the future value that results is S(t) = 1000e 0.10t $. What amount results in 5 years?
5.1 33) Drug in the bloodstream The percentage concentration y of a certain drug in the bloodstream at any time t in minutes is given by: y(t) = 100(1 e a) Which concentration results at 0 minutes? b) Which concentration results at 5 minutes? c) Which concentration results at 10 minutes? d) Interpret the results.
0.462t
).
5.1 45) Modelling Personal Income The table below depicts total personal income in USA (in billions of dollars) for selected years between 1960 to 2005. Year Personal Income
1960 411.5
1970 838.8
1980 2307.9
1990 4878.6
2000 8429.7
2005 10,239.2
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The data can be modelled as an exponential growth function. The equation of this function, with t as the number of years past 1960 is y(t) = 445.172(1.076 t ) (excel does only provide exponential functions with base e for this case). a) If the model is accurate, what will be the total U.S. personal income in 2010? b) In what year does the model predict total personal income will reach $22 trillion?
5.2 59) Decibel The loudness of sound (in decibels) perceived by the human ear depends on I intensity levels according to: L = 10 log with I0 = hearing threshold of an average human I0 ear. Find the loudness when I is 10,000 times I0 = intensity level of the average voice.
5.2 63), 65) pH Levels Chemists use the pH (hydrogen potential) of a solution to measure its + acidity of basicity. The pH is given by the formula: pH = log H with H + =
concentration of hydrogen ions in moles per liter. 63) Most common solutions have pH ranges between 1 and 14. What values of H + are associated with these extremes? 65) Sometimes pH is defined as the logarithm of the reciprocal of the concentration of hydrogen ions. Write an equation that represents this sentence, and explain how it and the equation given in the information in the equation can both represent pH.
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5.3 25) Inflation The purchasing power P (in $) of an annual amount of A $ after t years of 5% 0.05t
inflation follows the following function: P = Ae . a) How long will it be before a pension of $60,000 per year has a purchasing power of $30,000? b) How much pension A would be needed so that the purchasing power P is $50,000 after 15 years?
5.3 27) Radioactive half-life An initial amount of 100 g of the radioactive isotope thorium-234 decays according to: Q(t) = 100e 0.02828t , t in years. How long will it take until half of the initial amount is disintegrated (half-life of the isotope)?
100e q 5.3 35) Supply If the supply function for a product is given by p(q) = , q = number of q +1 units in 100. What will be the price when the producers are willing to supply 300 units?
5.3 37) Total cost The total cost function for a product is: C(x) = e 0.1x + 400 , x = number of items produced. What is the total cost of producing 30 units?
5.3 39) Total revenue The demand function for a product is: p(x) = 200e 0.02 x , p = price per unit when x units are demanded. What is the total revenue when 100 units are demanded?
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5.3 63) Mutual funds For annual data since 1978, the number of mutual funds N, excluding 8750 money market funds, can be modelled by: N(t) = , t = years past 1975. 1+ 80.8e 0.2286t a) Estimate the number of mutual funds in 2008. b) Estimate the year when the number of mutual funds will reach 8725. Solutions 5.1 32) S(5) = 1000e 0.10 5 = 1648.72 $ 5.1 33) a) y(t) = 100(1 e0 ) = 0 ; b) y(t) = 100(1 e 0.462 5 ) = 90.07% ; c) y(t) = 100(1 e 0.462 10 ) = 99.01% ; d) The first minutes the concentration raises rapidly, afterwards it rises less rapidly. 5.1 45) a) $17,645 billion / $17,343.6862 bn;
b) 2013
5.2 59) L = 10 log (10000 ) = 10 4 = 40 5.2 63) 0.1 and 1 10
14
1 H+
5.2 65) pH = log 5.3 25) a) 30 = 60e
0.05t
b) 50,000 = Ae 5.3 27) 50 = 100e
= log(1) log H + = log H + e 0.05t = 2
0.0515
0.02828t
ln(2) = 0.05t
t=
ln(2) = 13.86 years ; 0.05
A = 50,000 e 0.0515 = $105850.00 ln(2) e 0.02828t = 2 = t = 24.5 years . 0.02828
100e3 = $502.14 4 5.3 37) C(30) = e 0.1 30 + 400 = $420.09
5.3 35) p(3) =
5.3 39) p(100) = 200e
0.02 100
= $27.0671 = price for 100th unit. For 100 units this is: $2706.71. 8750 = 8391.08 8391 funds ; 5.3 63) a) N(33) = 1+ 80.8e 0.2286 33 8750 8750 80.8 8725 = 8725 (1+ 80.8e 0.2286t ) = 8750 1 = 0.0028 = 0.2286t 0.2286t 1+ 80.8e 8725 e b) 80.8 80.8 e 0.2286t = 0.2286t = ln t = 44.825 2019.8 in year 2019/2020 0.0028 0.0028
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4th Seminar Business Mathematics Systems of Linear Equations Self-Evaluation Exam I Additional Material: Harshbarger, Reynolds: Chapter 1, Section 1.5, Odd numbered questions from exercises of section 1.5 (Solutions at pp A9 – A10)
Concept of functions (pp 71) Relation: set of ordered pairs and rule how to find these pairs Function: relation between two or more sets. For each input element there is exactly one output element. Input: the independent variable. Output: the dependent variable. It is not a function, if more than one output element corresponds to one input element. Notation: y = f(x). f(x) corresponds to the rule how x is operated to receive a value of y and f(x) is what we see when we graph the function. ! Graphing points into a coordinate system ! Graphing a function using a value – table Domain: values of x, Range: values of y. The domains are restricted whenever values as 0 in the denominator or even roots of negative numbers result in the range (Example 7 p. 77). Operations with functions: sum, difference, product, quotient (denominator ≠ 0) are possible. Composite functions g of f (g f ) and f of g ( f g) → (g f )(x) = g( f (x)) and ( f g)(x) = f (g(x))
Basics of linear functions (pp 85) General form of a linear function: f (x) = mx + b . A linear function has a slope (a), an intercept with the y-Axis (y-Axis-Intercept = b), and with the x-Axis (null). Both intercepts lead to a b specific point of the linear function, namely: x = 0 ⇒ (0;b); y = 0 ⇒ − ;0 . These two points a allow drawing the graph of the linear function. y − y ∧y The slope m indicates the rate of change (also for non-linear functions). m = 2 1 = . x2 − x1 ∧x Directions of slopes: positive (m > 0), negative (m < 0), horizontal (m=0), vertical (m not 1 defined). Two lines l1 and l2 are perpendicular, if m1 = − . m2 General form: ax + by + c = 0; Point-slope form: y − y1 = m(x − x1 ); Slope-intercept form: y = mx + b Vertical line: x = a; Horizontal line: y = b
Breakeven Analysis (pp 119) C(x) = Total cost (fixed cost + variable cost) = cfix + cvar R(x) = Total revenue from sale of x units = (price per unit) • (number of units) P(x) = Profit from sale of x units P(x) = R(x) – C(x) Break-even point: cost = revenue: R(x) = C(x) Marginal profit (MP) = profit of one additional unit (slope of profit function) Marginal cost (MC) = cost to produce one additional unit (slope of cost function) Marginal revenue (MR) = revenue of selling one additional unit (slope of revenue function)
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Market Equilibrium Analysis (pp 122) At market equilibrium demand = supply (market cleared). Law of demand: as demand increases the price increases as well. Law of supply: as supply increases, the price decreases as well. For convenience, economists draw the quantity demanded or supplied (q) as the independent variable and the price (p) as the dependent. At market equilibrium, the equilibrium quantity and the equilibrium price apply. Demand and supply curves intersect. To find any equilibrium (BEP or market equilibrium), a system of equations has to be solved.
Parabolas (pp 151) General form: f (x) = ax 2 + bx + c For f (x) = ax 2 vertex a is turning point, minimal for a > 0, hence opens upward and maximal for a < 0, hence opens downward. b b Vertex point for f (x) = ax 2 + bx + c → ... → − ; f − setting y = 0 gives two null 2a 2a (intercepts). Because of symmetry the vertex has to be in the middle of these two points. Parabolas do not have “slopes” as linear functions, instead they have “average rate of change” as all non-linear functions. Similar to linear functions, the average rate of change can be found between two values of x, for example x1 = a, and x2 = b applying the “slope” formula: f (b) − f (a) . The secant connecting the two points (a; f(a)), and (b; Average rate of change = b−a f(b)) gets closer and closer to the actual function the nearer a is at b.
Business Applications of Quadratic Functions (pp 161) Quadratic or mixed supply and demand curves at 1st quadrant. Solving these equations leads to a system of quadratic equations. Similarly quadratic cost curves are used for break – even analysis. For monopoly markets: Revenue depends on price and demand ⇒ R(x) = p • x.
Exponential functions (pp 357) General form: f (x) = a x with a ≠ 0 The following functions are also exponential functions: f (x) = ca x ; f (x) = ca − x ; f (x) = ca bx Exponential growth function: f (x) = Ca x , y-intercept at (0, C), domain: x ∈ ℜ , range: y > 0. n
1 For continuously compounded interest e is used as basis as e = lim 1+ . For n being an n→∞ n infinitively small time step and infinitively many of these time steps. If 0 < a < 1, exponential decay applies (radioactive decay, decay of an advertising campaign, inflation on purchasing power effect).
Logarithmic functions (pp 369) Logarithmic functions solve problems where the unknown is the exponent. for y = log a (x) with a > 0 and a ≠ 1 (logarithmic form) ⇒ x = a y (exponential form) (logarithmic and exponential functions are inverse to each other, as one can be derived from the other by exchanging x and y and solve for y afterwards. Inverse functions have the same shape mirrored at the function y = x.
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Change of base of logarithmic functions can be used to simplify / solve. For example: log(x) ln(x) log b (x) = = log(b) ln(b)
Applications of Exponential and Logarithmic functions (pp 382) Solve exponential equations by using properties of logarithms. Isolate the exponential, take the logarithm on both sides, remove the variable from the exponent, solve for the variable.
Inverse Function The inverse of a function “mirrors” the original function at the line y(x) = x. It can be found by 1.) replacing x and y and 2.) solving for y. Example: y(x) = 3x + 5 1.) x = 3y + 5; 2.) y(x) = 0.33x -1.67 Some inverse functions and their original functions: f (x) = ln(x) ↔ f (x) = e x f (x) = log(x) ↔ f (x) = 10 x f (x) = sin(x) ↔ f (x) = arcsin(x)
Find the inverse of the following functions: f (x) = 2 x ↔
f (x) =
x +1 ↔ x
f (x) = x 2 + ax +
a2 ↔ 4
f (x) = 2 x ↔ f (x) = log 2 (x) = f (x) =
log(x) log(2)
x +1 y +1 1 ↔x= ⇒ xy = y +1 ⇒ y(x −1) = 1 ⇒ f (x) = x y x −1 2
2
a a a2 a a f (x) = x + ax + = x + ↔ x = y + ⇒ ± x = y + ⇒ f (x) = ± x − 2 2 4 2 2 2
Self-Evaluation exam I Time allowed: 20 Minutes Aids allowed: Non-programmable non-graphical calculator
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Graphical solution of a system of linear equations Draw the linear equations as functions and find their intersection point. Example 1 I)
2x – y = 0
II)
2x + y = 8
Solve both equations for y: I)
y = 2x
II)
y = - 2x + 8
Graph both equations as if these were functions:
Find out their intersection point by looking at the graph. It looks like it is (2; 4). Now solve the system by computing their intersection point: Set I) = II) as it is already solved for y: I)
y = 2x = II)
y = - 2x + 8
2x = - 2x + 8 " 4x = 8 and x = 2. Now plug in the solution for x into I) or II): I)
y = 2 • 2 = 4 or into II)
y = - 2• 2 + 8 = 4 " y = 4 " (2; 4) is the right solution.
An alternative way to solve that system is the addition method: I)
2x – y = 0
II)
2x + y = 8
Add both equations together such that one of the variables is removed. In our case - y in equation I) and +y in equation II) are eliminated when we add both equations together.
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2x – y = 0
II)
2x + y = 8
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"
I) + II) 4x = 8
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x=2
Again plug in the solution for x into I) or II) to get the solution for y: "
I) 2 • 2 – y = 0
y=4 "
Solution: (2; 4)
Both ways provide the same (correct) solution. Sometimes it is just quicker or easier to use the one or the other method.
Calculated (analytical) solution of a system of linear equations Solve the system of linear equations by calculating the intersection point. Example 2 I)
4x + 5y = 18
II)
3x – 9y = – 12
If we add these two equations neither x nor y would disappear, therefore we first have to manipulate them. To eliminate x equation I) is multiplied by 3 and equation II) by – 4. 3 • I)
12x + 15y = 54
(-4) • II)
–12x + 36y = 48
Adding the two equations provide one equation with only one variable: y. 3 • I)
12x + 15y = 54
(-4) • II)
–12x + 36y = 48 51y = 102
"
y=2
Plug into equation I) or II) gives: I)
4x + 5 • 2 = 18
"
4x = 8 "
x=2 "
Solution: (2; 2)
To check, if that solution is correct, plug both values into equation II) II)
3•2 – 9•2 = – 12
(correct)
Text problems Example 3: Investment mix (p 110) A person has $200,000 invested, part at 9% and part at 8%. If the total yearly income from the two investments is $17,200, how much is invested at 9% and how much at 8%? This problem represents a system of two equations, but first the variables must be declared: x = amount invested at 9%, y = amount invested at 8%, x + y = total investment I) II)
x + y = 200,000 0.09x + 0.08y = 17,200
Multiply equation I) by - 8 and equation II) by 100 to remove y. I)
- 8x + - 8y
= - 1,600,000
II)
9x + 8y
= 1,720,000
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x = 120,000
Plugging into equation I) gives: I)
120,000 + y
= 200,000
" y = 80,000
Solution: $120,000 is invested at 9% and $80,000 is invested at 8%. Check with equation II) gives: II)
0.09 • 120,000 + 0.08 • 80,000 = 10,800 + 6,400 = 17,200 (correct)
Systems with no or infinite solutions If two equations do not provide a solution, these two equations do not intersect, they are parallel. If two equations provide an infinite number of solutions, both equations are “dependent” or could be manipulated to become identical. Example 4: No intersection (p 111) I)
4x + 3y = 4
II)
8x + 6y = 18
(-2) • I)
-8x - 6y = -8
Adding I) + II) II)
8x + 6y = 18
(-2) • I)
-8x - 6y = -8 0 = 10
As this is a contradiction, these two equations do not intersect, hence they are parallel. Example 5: Infinite number of intersections I)
4x + 3y = 4
II)
8x + 6y = 8
(-2) • I)
-8x - 6y = -8
Adding I) + II) II)
8x + 6y = 8
(-2) • I)
-8x - 6y = -8 0=0
This is an identity, therefore their graphs coincide, and each point in the graph is a solution.
Three equations with three unknowns The same methods used before can also be used for more than 2 unknowns. Graphically there would be a coordinate system with x, y, and z axis. Example 6: I)
x+y+z
=6
II)
3x + y
=5
III)
y-z
=-1
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First one variable is eliminated to receive a system of two equations with two unknowns. Then this system is solved as before. Eliminating z by adding equations I) and III) together I)
x+y+z
III)
y-z
1)
x + 2y
=6 =-1 =5
Now solving the system of the two equations 1) and II) 1)
x + 2y
=5
II)
3x + y
=5
Eliminating y by multiplying equation II) by (-2) and adding both equations together 1)
x + 2y
=5
(-2) • II)
-6x - 2y
= -10
- 5x
=-5
"x=1
Plugging into equation 1) gives: 1)
1 + 2y
=5
"y=2
Plugging values of x and y into equation I) gives I)
1+2+z
=6
"z=3
Control by plugging into equation III) III)
2-3
= - 1 (correct)
In class examples 1) Solve the system of three equations with three unknowns. I)
3x + 2y + z
=6
II)
x–y–z
=0
III)
x+y–z
=4
Solution:
x = 1, y = 2, z = -1
2) Solve the system of three equations with three unknowns. I) II) III)
x-y-z y – 2z x+y+z
=0 = - 18 =6
Solution:
x = 3, y = -4, z = 7
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5th Seminar Business Mathematics Systems of Linear Equations Revisitation: Basics of Algebra From textbook: Kaufmann, J.E. und Schwitters, K.L. (2009) College Algebra, 7th International Edition, Brooks/Cole; ISBN-13:978-0-495-55403-5
Systems of simultaneous linear equations with two variables (6.1 p 481 – 494) Graph of two linear equations in two variables is two straight lines, which can cross (slopes not equal, one solution), are parallel to each other (equal slopes, no solution), or identical to each other (infinite number of solutions).
x+y=6 Examples: 1) x−y=2
3x + 2y = 1 2) 5x − 2y = 23
4x − 5y = 21 3) −3x + y = −7
Solution = Intersection of the two straight lines. Solution 1) {4; 2} Solution check by substitution of x and y through the values 4 and 2 gives: 1) x + y = 6 = 4 + 2 (true) and x − y = 2 = 4 − 2 (true) . Two methods to solve systems of equations: 1) substitution method and 2) elimination by addition method (Gauss method). 1) Solve for one of the variables and substitute it in the other equation 2) Add, multiply, subtract and change equation order to simplify the system of equations. Proceed until one of the variables can be eliminated by adding or subtracting equations. Examples
1)
x − 3y = −25 4x + 5y = 19
=?
2)
3x + 5y = −9 2x − 3y = 13
=?
3)
4)
x − 4y = 9 2x − 8y = 7
5)
1 2 x + y = −4 2 3 1 3 x − y = 20 4 2
=?
=?
BJ runs his boat at full throttle, which results in the boat traveling at a constant speed. Going up the river against the current, the boat travelled 72 miles in 4.5 hours. The
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6)
7)
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return trip down the river with the current only took three hours. Find the speed of the boat and the speed of the current. Lucinda invested $950, part of it at 6% interest and the remainder at 8%. Her total yearly income from the two investments was $71.00. How much did she invest at which rate? The units digit of a two-digit number is one more than twice the tens digit. The number with the digits reversed is 45 larger than the original number. Find the original number.
Solutions from examples above
x − 3y = −25 ⇒ x = 3y − 25 1) 4x + 5y = 19 ⇒ 4(3y − 25) + 5y = 19
12y −100 + 5y = 19 ⇒ 17y = 119 ⇒ y = 7 set y = 7 in x = 3y − 25 ⇒ x = 21− 25 = −4 ⇒ {−4; 7}
3x + 5y = −9 ⋅2 2)
⇒
2x − 3y = 13 ⋅(−3) ⇒
6x +10y = −18 −6x + 9y = −39 ⇒ 2x = 3y +13 = 3(−3) +13 = 4 ⇒ x = 2 19y = −57 ⇒ y = −3
1 2 x + y = −4 ⋅6 ⇒ 3x + 4y = −24 2 3 3) 1 3 −3x +18y = −240 x − y = 20 ⋅4 ⇒ x − 6y = 80 ⋅(−3) ⇒ ⇒ y = −12 ⇒ x = 8 4 2 22y = −264
{2;−3}
{8;−12}
x − 4y = 9 ⋅(−2) ⇒ − 2x + 8y = −18 4)
2x − 8y = 7 ⇒ contradiction {∅} 0 + 0 = −11
5) x: speed of boat, y: speed of current. Against current: 72 miles/4.5 hours = 16 miles per hour. With the current: 72 miles / 3 hours = 24 miles per hour. x + y = 24
x − y = 16 ⇒ x = 20 ⇒ y = 24 − 20 = 4 {20 = speed of boat ; 4 = speed of current} 2x = 40 6) x: amount invested at 6%, y: amount invested at 8%. x + y = 950 ⋅(−6) ⇒ −6x − 6y = −5700 6x + 8y = 7100 2y = 1400
0.06x + 0.08y = 71 ⋅100 ⇒ y = 700; x = 950 − 700 = 250
{250$ = investment at 6% ; 700$ = investment at 8%}
7) u: units digit, t: tens digit. 10t+u represents original two-digit number. u = 2t + 1 10u + t = 10t + u + 45 ⇔ 9u − 9t = 45 ⇔ u − t = 5
substitute: 2t + 1− t = 5 ⇒ t = 4 and u = 2 ⋅ 4 + 1 = 9 ⇒
{4 = tens digit;9 =
units digit solution: 49}
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Examples from the book
1) 3) 13) 9) 23) 29)
x + y = 16 =? y = x + 2 x = 3y − 25 =? 4x + 5y = 19 4x + 3y = −7 =? 3x − 2y = 16 2x − 3y = 4 2 4 =? y= x− 3 3 4x − 5y = 21 =? 3x + 7y = −38 2 1 s + t = −1 3 4 =? 1 1 s − t = −7 2 3
Solution: {7; 9}
Solution: {- 4; 7}
Solution: {2; - 5}
2 4 Solution: k; k − 3 3 Solution: {- 1; - 5}
Solution: {s = - 6; t = 12}
65) One day last summer Jim went kayaking on the Little Susitna River in Alaska. Paddling upstream against the current, he travelled 20 miles in 4 hours. Then he turned around and paddled twice as fast downstream and, with the help of the current, travelled 19 miles in 1 hour. Find the rate of the current. Solution: 3 mph 63) Sam invested $1950, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $6 more than twice the income from the 6% investment. How much did he invest at each rate? Solution: $750 at 6%, $1200 at 8% 57) The sum of the digits of a two-digit number is 7. If the digits are reversed, the newly formed number is 9 larger than the original number. Find the original number. Solution: 34
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Revisitation: Basics of Algebra Properties of real numbers Closure property:
a + b is a unique real number
Commutative property:
a + b = b + a,
and
ab = ba
Associative property:
(a + b) + c = a + (b + c),
and
(ab)c = a(bc)
Identity property:
a + 0 = 0 + a = a,
and
a(1) = 1(a) = a
Inverse property:
a + (–a) = (–a) + a = 0,
and
a(1/a) = (1/a)a = 1
Multiplication property of 0:
a(0) = (0)a = 0
Multiplication property of (-1):
a(-1) = –1(a) = – a
Distributive property:
a(b + c) = ab + ac
Exponents (0.2 p 18 – 30) x2: x = base, 2 = exponent; xm = x · x · x · …. · x (m times x multiplied with itself) Properties of exponents xm • xn = xm + n (xm)n = xmn xn • yn = (xy)n n x xn = n; y≠0 y y
xm = x m−n ; x ≠ 0 xn xm + 3xm = 4xm
but
xm + x2m ≠ 2x3m, also not 2xm, or 2x2m, it just remains
xm + ym ≠ (x + y)m
but
(x2 – y2) = (x + y) (x – y)
x 5m = x 5m−2m = x 3m 2m x
note: x0 = 1,
m 2
2 m
(x ) = (x ) x −m =
1 xm
(3rd binomial formula)
= x 2m
1 = xm −m x
Roots (0.6 p 64 – 75) If the exponent is not an integer, the exponent properties above apply. Write roots in exponent form. E. g. √a = a1/2 , etc. Note: the root of a negative expression is not a real number. Note: √25 = ±5 because 52 =25 and (-5)2 =25, hence always two solutions. Positive real numbers have two square roots, but negative numbers don’t because the square of any real number is always positive (compare above). Cube roots of negative real numbers give real numbers, because: (-2)3 = – 8, so the reverse calculation gives: 3 −8 = −2 .
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Overview Real number Positive Root (n) Even Odd
n
Negative
Two Solutions (± a) One real positive cube root, two imaginary
b =a
if and only if
an = b
and
( b) = b n
Properties of roots (similar to exponents) n
bc = n b ⋅ n c if b, c ∈ ℜ
n
b nb = c nc
if b, c ∈ ℜ and c ≠ 0
b m−n = b b
if b ∈ ℜ and b ≠ 0
m n m
b ⋅ n b = m+n b
m n
b = m⋅n b
Exponents and Roots (0.7 p 75 – 82) m n
b = n b m = ( n b )m
Examples 2
1)
83 = ?
2)
x2 ⋅ x3 = ?
3)
2 2 16y 3 = ?
1
2 1
4)
y y
3 4 1 2
=? 6
5)
1 x2 1 = ? y3
xy ⋅ 5 x 2 y = ?
6) 7) 8)
3
x2 ⋅ 4 x3 = ?
Rationalise denominator
2 =? x
3
Solutions from examples above 1)
2 3
8 = 3 82 = 3 64 = 4 try with calculator: 8^(2/3)
No real solution (imaginary) One real negative cube root, two imaginary
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1
Page 6 1 2 + 3
3 4 + 6
7
2)
x2 ⋅ x3 = x2
3)
2 2 2 1 1 1 3 3 2 2 16y = 16 (y ) = 4y 3
5thSem14.doc
1
= x6 = x ⋅ x6 = x ⋅ 6 x
= x6
1
3
4)
y4 y
1 2
=y
3 1 − 4 2
=y
1 4
6
5)
1 x2 1 y3
6
1 x2 x3 = = 6 y2 1 3 y 1
6)
2
7)
3
1
2 3 + 4
3
x2 ⋅ 4 x3 = x 3 ⋅ x 4 = x 3 2
8)
2
1
1 2 +
1 1 + 5
xy ⋅ 5 x 2 y = x 2 y 2 x 5 y 5 = x 2 5 y 2 8
= x 12
9 + 12
= x 0.9 y 0.7 17
5
= x 12 = x ⋅ x 12 = x ⋅ 12 x 5
2
2 2 x 3 2x 3 = ⋅ 2 = 1 3 x x 3 x x3
Logarithms (5.3 p 439 – 461) log b (r) = t
⇔
bt = r
log 2 (8) = 3
⇔
23 = 8
log10 (100) = 2
⇔
10 2 = 100
For b > 0, b ≠ 1 For b > 0, b ≠ 1, r > 0 For b,r,s > 0, b,r,s ≠ 1 For b,r,s > 0, b,r,s ≠ 1
log b (b) = 1, log b (1) = 0 b logb (r ) = r log b (rs) = log b (r) + log b (s) r log b = log b (r) − log b (s) s
log b (r b ) = b ⋅ log b (r) log10 (a) log c (a) = Change of basis of logarithms: log b (a) = log10 (b) log c (b) Examples For r,b > 0, b ≠ 1
1) 2) 3) 4) 5)
log10 (0.0001) = x, x = ? 5 27 log 9 =? 3 2 log8 ( x ) = x=? 3 27 log b = 3 b = ? 64 log 2 ( 5) = 2.3219, log 2 (3) = 1.5850
log 2 (15) = ?
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6)
log 5 (36 ) = 2.2266, log 5 ( 4) = 0.8614
7)
1 log 2 22 3 = ?
8)
Solve the brackets
x log b =? yz
9)
Write as one expression
2 logb ( x) − 3 logb ( y) − 4 logb ( z) = ?
10)
log 5 ( x + 4) − log 5 (x) = 2
x =?
11)
log10 ( x ) + log10 (x + 9) = 1
12)
log 2 (3) + log 2 (x + 4) = 3
log 5 ( 9 ) = ?
x =? x =?
Solutions from examples above 1)
log10 (0.0001) ⇔ 10 x = 0.0001 ⇒ 10 x = 10 −4 ⇒ x = −4
2)
2 5 27 − 27 5 35 2 1 log 9 ⇒ 32 x = = 3 5 ⇒ 2x = − ⇒ x = − = x ⇒ 9x = 3 3 5 5 3
3)
2 log8 ( x ) = 3
4)
27 27 27 3 27 3 log b = 3 ⇒ b3 = ⇒b= 3 = = 64 64 64 3 64 4 log 2 (15) = log 2 ( 5⋅ 3) = log 2 ( 5) + log 2 (3) = 2.3219 +1.5850 = 3.9069
1
5) 6) 7)
3
2
1 ⇒ x = 8 = 8 3 = 2 2 = 4 2 3
36 log 5 ( 9 ) = log 5 = log 5 (36) − log 5 (4) = 2.2266 − 0.8614 = 1.3652 4 1 1 log10 (22) 1 log 2 22 3 = ⋅ log 2 (22) = = ⋅ 4.4594 = 1.4865 3⋅ log10 (2) 3 3 1
8) 9) 10)
11)
12)
x x 2 1 log b = log b = ⋅ ( log b (x) − log b (y) − log b (z)) 2 yz yz x2 2 log b (x) + 3log b (y) − 4 log b (z) = log b (x 2 ) + log b (y 3 ) − log b (z 4 ) = log b 3 4 y z x + 4 x+4 1 log 5 = 52 ⇒ 24x = 4 ⇒ x = =2 ⇒ x x 6
log10 ( x ) + log10 (x + 9) = 1 ⇒ log10 [ x(x + 9)] = 1 ⇒ x(x + 9) = 101 x 2 + 9x −10 = 0 = (x +10)(x −1) ⇒ x1 = −10, x2 = 1
Only x2 is a solution!
log 2 (3) + log 2 (x + 4) = 3 ⇒ log 2 [3⋅ (x + 4)] = 3 ⇒ [3⋅ (x + 4)] = 2 3 3x +12 = 8 ⇒ 3x = −4 ⇒ x = −
4 3
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Multiplication of polynomials With monomial
3x 2 ⋅ (2x 2 + 5x + 3) = 6x 4 +15x 3 + 9x 2
Binomial
( x + 2) ⋅ (y + 5) = xy + 5x + 2y +10
(x − 3)⋅ ( y + z + 3) = x ⋅ ( y + z + 3) − 3 ⋅ ( y + z + 3) = xy + xz + 3x − 3y − 3z − 9
Trinomial
Multiply and simplify
( x + 5) ⋅ (x + 7) = x 2 + 5x + 7x + 35 = x 2 +12x + 35 (2x + 5) ⋅ (3x − 2) = 6x 2 +15x − 4x −10 = 6x 2 +11x −10
Check correctness of a solution by inserting a number (for example x = 4)
( x − 2) ⋅ (x 2 − 3x + 4) = x ⋅ (x 2 − 3x + 4) − 2 ⋅ (x 2 − 3x + 4) = x 3 − 3x 2 + 4x − 2x 2 + 6x − 8 = x 3 − 5x 2 +10x − 8 ( x − 2) ⋅ (x 2 − 3x + 4) = x 3 − 5x 2 +10x − 8 ( 4 − 2) ⋅ (4 2 − 3⋅ 4 + 4) = 43 − 5⋅ 4 2 +10 ⋅ 4 − 8 2 ⋅ (16 −12 + 4) = 16 = 64 − 80 + 40 − 8 Special patterns (binomial expansion pattern, general pattern) 1
( a ± b) = a ± b 2 ( a ± b) = a 2 ± 2ab + b 2 3 ( a − b) = a3 − 3a 2 b + 3ab 2 − b3 4 ( a − b) = a 4 − 4a3b + 6a 2 b 2 − 4ab3 + b 4 5 ( a − b) = a 5 − 5a 4 b +10a3b 2 −10a 2 b3 + 5ab 4 − b 5 Factoring polynomials (0.4 p 40-49) Examples 1)
Expand (a − b)4 = ?
2)
Expand (2x + 3y)5 = ?
Solutions from examples above 1) 2)
( a − b)
4
= a 4 − 4a 3b + 6a 2 b 2 − 4ab3 + b 4
(2x + 3y)5 = (2x)5 + 5⋅ (2x)4 ⋅ (3y) +10 ⋅ (2x)3 ⋅ (3y)2 +10 ⋅ (2x)2 ⋅ (3y)3 + 5⋅ (2x)⋅ (3y)4 + (3y)5 = 32x 5 + 240x 4 y + 720x 3 y 2 +1080x 2 y 3 + 810xy 4 + 243y 5
Examples from the book 57) (x − y)5 = ?
Solution: x 5 − 5x 4 y +10x 3 y 2 −10x 2 y 3 + 5xy 4 − y 5
65) (2a − 3b)5 = ?
Solution: 32a 5 − 240a 4b + 720a 3b 2 − 1080a 2b 3 + 810ab 4 − 243b 5
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Solving quadratic equations (1.3 p 120 – 133) Quadratic equations have the form: a1 x 2 + a2 x + b = 0 or ax 2 + bx + c = 0 For example: x 2 + 2x −15 = 0 If the equation is factorable the solution can be found immediately: x 2 + 2x −15 = (x + 5)(x − 3) = 0 ⇒ x + 5 = 0 and/or x − 3 = 0 ⇒ x1 = −5, x2 = 3 {5;3} −b ± b 2 − 4ac b c [2] x1 + x2 = − ; x1 x2 = 2a a a Quadratic equations can have either two, one double or complex solutions.
Quadratic formulas: ax 2 + bx + c = 0 ⇒ [1] x1,2 =
Examples 1)
n = −6n 2 +12
2)
x2 = k
3)
x 2 = 72
4)
(3n −1)2 = 26
5)
x 2 + 8x − 2 = 0 Build a binomial to solve the equation
6)
2x 2 + 6x − 3 = 0
7)
3x 2 − x − 5 = 0
8)
25n 2 − 30n = −9
9)
x 2 − 4x −192 = 0 Use formulas [2] to check the solution
10)
One leg of a right triangle is 7 meters longer than the other leg. If the length of the
n=?
x =? k ∈ ℜ x =?
hypotenuse is 17 meters, find the length of each leg. Solutions from examples above
4 3 ;− 3 2
1)
4 3 6n 2 + n −12 = 0 = (3n − 4)(2n + 3) ⇒ n1 = ;n2 = − 3 2
2)
x 2 = k ⇒ x 2 − k = 0 = ( x + k )( x − k ) ⇒ x1 = − k x2 = k
3)
x 2 = 72 ⇒ x1 = ± 72 = ±6 2
4)
3n −1 = ± 26 ⇒ n1,2 =
1± 26 3
{±6 2 } 1± 26 3
{± k }
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x 2 + 8x − 2 = 0 → x 2 + 8x = 2 +16 x 2 + 8x +16 = 18 ⇒ (x + 4)2 = 18 ⇒ x + 4 = ± 18 = ±3 2 ⇒ x1,2 = −4 ± 3 2 2
6)
2x 2 + 6x = 3 ÷2 → x 2 + 3x =
3 3 3 9 15 3 15 → x + = + = ⇒ x + = ± 2 2 2 4 4 2 4
−(−1) ± (−1)2 − 4(3)(−5) 1± 61 = 2(3) 6
7)
3x 2 − x − 5 = 0 ⇒ x1,2 =
8)
25n 2 − 30n = −9 ⇒ n1,2 =
3± 15 2
1± 61 6
−(−30) ± (−30)2 − 4(25)(−9) 30 ± 0 = 2(25) 50
x 2 − 4x = 192 ⇒ (x − 2)2 = 192 + 4 ⇒ x − 2 = ±14
{−4 ± 3 2 }
3 5
{16;−12}
9)
b (−4) c −192 x1 + x2 = − ⇒ 16 −12 = − and x1 x2 = ⇒ (16)(−12) = a 1 a 1
10)
l 2 + (l + 7)2 = 172 ⇒ 2l 2 +14l − 240 = 0 = (l +15)(l − 8) ⇒ l1,2 = (−15);8 {l1 = 8;l2 = 8 + 7 = 15}
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6th Seminar Business Mathematics Sequences, simple and compound interest calculations Additional Material: Harshbarger, Reynolds: Chapter 6, Sections 6.1, 6.2. Odd numbered questions from exercises of sections 6.1, and 6.2 (Solutions at pp A33 – A34)
Simple Interest; Sequences (pp 404) Simple Interest I from a sum of money P (principal) is given by: I = P ⋅ r ⋅ t r : annual interest rate (as decimal); t: time (in years) Simple interest just “adds” the interest rate to the principal money each year. It disregards the “interest of the interest” rate. Future value of investment or loan: S = P + I Example 1: Future and present value (p 405) a) If $2,000 is borrowed for on-half year at a simple interest rate of 12% per year, what is the future value of the loan at the end of the half-year? b) An investor wants to have $20,000 in 9 months. If the best available interest rate is 6.05% per year, how much must be invested to yield the desired amount?
Example 2: Return on Investment (pp 405) Ms Spaulding bought Wind-Gen Electric stock for $6,125.00 and after 6 months, the value of her shares had risen by $128.00 and dividends totalling $144.14 had been paid. Find the simple interest rate she earned on this investment if she sold the stock at the end of the 6 months.
Example 3: Duration of an Investment (p 406) If $1,000 is invested at 5.8% simple interest, how long will it take to grow to $1,100?
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Sequences (pp 406) Sequence function: domain is the set of positive integers. The outputs of the sequence function form an ordered list called “sequence” sn, which is the sum of the terms a1, a2, a3, … , an. Example 4: Terms of a sequence (p 407) a) Write the first four terms of the sequence whose nth term is an = b) Write the sequence s4.
(−1)n 2n
Arithmetic Sequences (pp 407) Arithmetic sequences have common differences between two terms: an = an−1 + d for n > 1 . As all differences are d, we can compute an, if a1 and d are given: an = a1 + (n −1)⋅ d .
Sn = a1 + (a1 + d) + (a1 + 2d) + ...+ [ a1 + (n −1)d ] Sum of first n terms of an arithmetic sequence: Sn = an + (an − d) + (an − 2d) + ...+ [ an − (n −1)d ]
n 2Sn = n ⋅ (a1 + an ) ⇒ Sn = ⋅ (a1 + an ) 2 Example 5: nth term and sum of an arithmetic sequence (pp 408) a) Find the 12th term of the arithmetic sequence with first term = 3 and common difference = 2. b) Find the sum of the first 12 terms of this sequence.
Compound Interest; Geometric Sequences (pp 412) Compound interest adds the interest of each period to the principal.
r r i = ; n = mt; S = P ⋅ (1+ i)n = P ⋅ 1+ m m
mt
For continuous compounding this formula applies: S = P ⋅ ert Euler's number: e = 2.7182818... Example 6: How much more will you earn if you invest $1,000 for 5 years at 8% compounded continuously instead of at 8% compounded quarterly? (pp 417)
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Annual Percentage Yield (APY) interest earned from investing money compounded m r periodical: APY = 1+ −1 = (1+ i)m −1 , simple: APY = mi , or continuous: APY = er −1 . m Example 7: A young couple found three different investment companies that offered college savings plans: (a) one at 10% compounded annually, (b) another at 9.8% compounded quarterly, and (c) a third at 9.65% compounded continuously. Find the APY for each of these three plans in order to discover which plan is best (p 419).
Example 8: How long does it take an investment of $10,000 to double if it is invested at: 8a) 8% compounded annually? 8b) 8% compounded continuously? (p 419)
A geometric sequence has a common ratio: an = r ⋅ an−1 for n > 0 . The nth term of a geometric sequence is given by: an = a1 ⋅ r n−1 . Geometric sequences apply for example for compound interest rates. The sum of the first n terms of a geometric sequence:
Sn = a1 + a1 ⋅ r + a1 ⋅ r 2 + a1 ⋅ r 3 + ...+ a1 ⋅ r n−1 = a1 ⋅ (1+ r + r 2 + r 3 + ...+ r n−1 ) r ⋅ Sn = a1 ⋅ (r + r 2 + r 3 + r 4 + ...+ r n ) n
Sn − r ⋅ Sn = a1 ⋅ (1− r ) ⇒ Sn =
.
a1 ⋅ (1− r n )
with r ≠ 1 1− r Example 9: Find the seventh term of the geometric sequence with the first term = 5 and common ratio = - 2 (p 421).
Example 10: Find the sum of the first five terms of the geometric sequence with first term = 4 and common ratio = -3.
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Solutions from in class examples 1a) I = $2,000 • 0.12 • 0.5 = $120
S $20,000 ⇒P= = $19,131.89 1+ r ⋅ t 1+ 0.0605⋅ 0.75
1b)
S = P + I = P + P⋅r⋅t ⇒ P =
2.
Principal P = $6,125.00, time t = 0.5 years, Interest I = $128.00 + $144.14 = $272.14
I = P ⋅ r ⋅ t ⇒ $282.14 = $6,125⋅ r ⋅ 0.5 ⇒ r =
$272.14 = 0.089 = 8.9% $6,125⋅ 0.5 S − P $100 = = 1.72 years P⋅r $58
3.
P = $1,000,S = $1,100,r = 0.058 ⇒ S = P + P ⋅ r ⋅ t ⇒ t =
4a)
(−1)1 1 (−1)2 1 (−1)3 1 (−1)4 1 a1 = = − ; a2 = = ; a3 = = − ; a4 = = 2 2 2⋅2 4 2⋅3 6 2⋅4 8
4b)
1 1 1 1 −12 + 6 − 4 + 3 −7 S4 = a1 + a2 + a3 + a4 = − + − + = = = −0.29 2 4 6 8 24 24
5a)
a12 = 3+ (12 −1)⋅ 2 = 25
5b)
S12 =
6.
Continuously: S = $1,000 ⋅ e 0.08⋅5 = $1, 491.82; Quarterly: S = $1,000 ⋅ (1.02)20 = $1, 485.95 Extra interest for compounding continuously: $1, 491.82 − $1, 485.95 = $5.87
7a)
APY = (1+ 0.1)1 −1 →10%
7b)
0.098 APY = 1+ −1 = 1.10166 −1 →10.166% Best choice 4
7c)
APY = e 0.0965 −1 = 1.10131−1 →10.131% still better than a)
8a)
$20,000 = 10,000 ⋅ (1+ 0.08)n ⇒ 2 = 1.08n ⇒ n =
8b)
$20,000 = 10,000 ⋅ e 0.08t ⇒ 2 = e 0.08t ⇒ n =
9.
a7 = 5⋅ (−2)7−1 = 320
12 ⋅ (3+ 25) = 168 2
4
10.
S5 =
4 ⋅ (1− (−3)5 ) 1− (−3)
= 244
log(2) ≈ 9 years log(1.08)
ln(2) ≈ 8.7 years 0.08
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Applications in class (pp 242, pp 256) 6.1 11) Borrow money: If you borrow $800 for 6 months at 16% annual simple interest, how much must you repay at the end of the 6 months?
6.1 19) File cabinets: A firm buys 12 file cabinets at $140 each, with the bill due in 90 days. How much must the firm deposit now to have enough to pay the bill if the money is worth 12% per year? Use 360 days in a year.
6.1 32) Sequences: Write the first five terms of the sequence whose nth term is: an =
6.1 34) Sequences: Write the sixth term of the sequence whose nth term is: an =
(−1)n n2
n(n −1) n+3
6.1 52) Arithmetic Sequences: If you make $36,000 and get $2,400 raises each year, in how many years will your salary double?
6.1 56) Pay raises: As an employee, would you prefer being given a $1,200 raise each year for 5 years or a $200 raise each quarter for 5 years?
6.2 9) Future value: What is the future value if $8,000 is invested for 10 years at 12% compounded annually?
6.2 15) Lump sum: Which lump sum do parents need to deposit in an account earning 10%, compounded monthly, so that it will grow to $80,000 for their son’s college tuition in 18 years? 6.2 25) Investments: Which investment will earn more money, a $1,000 investment for 5 years at 8% compounded annually or the investment for 5 years compounded continuously at 7%? 6.2 46) Investments: How long does it take for an account containing $8,000 to be worth $15,000 if the money is invested at 9% compounded monthly?
6.2 63) Inflation: A house that 20 years ago was worth $160,000 has increased in value by 4% each year. What is its worth today? Hint: treat the 4% as an annual interest rate.
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Solutions 6.1 11) S = P + P ⋅ r ⋅ t = 800(1+ 0.16 ⋅ 0.5) = $864
90 $1,680 = $1,631.07 6.1 19) S = 12 ⋅ $140 = $1,680 = P 1+ 0.12 ⋅ = 1.03P ⇒ P = 360 1.03 6.1 32) a1 =
(−1)1 (−1)2 1 (−1)3 1 (−1)2 1 (−1)5 1 = −1; a = = ; a = = − ; a = = ; a = =− 2 3 4 5 2 2 2 2 2 1 2 4 3 9 4 16 5 25
6.1 34) a6 =
6 ⋅ (6 −1) 30 = = 3.33 6+3 9
an = a1 + (n −1)⋅ d ⇒ n = 6.1 52)
an − a1 +1 d
d = $2, 400; a1 = $38, 400; an = $72,000 ⇒ n = 1) a 5 = $1200 + 4 ⋅ $1,200 = $6,000;
6.1 56)
$33,600 +1 = 15 $2, 400
2) a20 = 200 +19 ⋅ 200 = $4,000
5 20 1) S5 = ($1200 + $6000) = $18,000; 2) S20 = ($200 + $4000) = $42,000 2 2
6.2 9) S = P ⋅ (1+ i)n = $8,000 ⋅ (1+ 0.12)10 = $24,846.79 12⋅18
0.1 6.2 15) $80,000 = P ⋅ 1+ 12
6.2 25)
⇒P=
$80,000 = $13,322.91 12⋅18 0.1 1+ 12
Sannual = P ⋅ (1+ i)n = $1,000 ⋅ (1+ 0.08)5 = $1, 469.33; Scontinuous = P ⋅ ert = $1,000 ⋅ e 0.07⋅5 = $1, 419.06 ⇒ $50.26 more annual compound at 8% 12⋅n
6.2 46) $15,000 = $8,000 ⋅ 1+ 0.09
12
15 log(1.875) ⇒ log = 12 ⋅ n ⋅ log(1.0075) ⇒ n = = 7.01 years 8 12 ⋅ log(1.0075)
6.2 63) S = $160,000 ⋅ (1+ 0.04)20 = $350,579.97
Self-Evaluation Exam • Systems of Linear Equations • Simple and Compound Interest Calculations
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7th Seminar Business Mathematics Future value of annuities Present value of annuities Self-Evaluation Exam II Additional Material: Harshbarger, Reynolds: Chapter 6, Sections 6.3, 6.4. Odd numbered questions from exercises of sections 6.3, and 6.4 (Solutions at pp A34 – A35)
Future value of annuities (pp 427) Annuities: an annual amount of money deposited with a certain interest rate for later use. When you deposit the same amount of money regularly at the same interest, this follows the law of a (1+ i)n −1 geometric sequence. S = R ⋅ ; S: future value of annuities; R: regularly deposited i (1+ i)n −1 amount; i: interest rate. s¬ = = future value, if $1 is deposited. ni i Example 1: Future value (pp 428) Richard deposits $200 at the end of each quarter in an account that pays 4%, compounded quarterly. How much money will he have in 2¼ years?
Example 2: Compare two Investments (pp 429) 2a) Twin 1 invests $2,000 at the end of each year for 8 years in an account that pays 10%, compounded annually. After the initial 8 years, no additional contributions are made, but the investment continues to earn 10%, compounded annually, for 36 more years. How much does twin 1 have at age 65? 2b) Twin 2 invests $2,000 at the end of each year for 36 years in an account that pays 10%, compounded annually. Twin 2 starts 8 years later than twin 1. How much does twin 1 have at age 65?
Example 3: Time to reach a financial goal (pp 430) A small business invests $1,000 at the end of each month in an account that earns 6% compounded monthly. How long will it take until the business has $100,000 toward the purchase of its own office building? (hint: use annuity formula and solve for “n”).
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Example 4: Amount of annuity (pp 431) A company establishes a sinking fund to discharge a dept of $300,000 due in 5 years by making equal semi-annual deposits, the first due in 6 months. If the deposits are placed in an account that pays 6%, compounded semi-annually, what is the size of the deposits? (hint: use annuity formula and solve for R).
Annuities due: periodic payments made at the beginning of each period, hence, interest for one (1+ i)n −1 more period Sdue = R ⋅ ⋅ (1+ i) . Solve for n, if time is asked (n: number of periods): i
S ⋅i log due +1 (1+ i) −1 Sdue ⋅ i R ⋅ (1+ i) Sdue = R ⋅ +1 = (1+ i)n ⇒ n = ⋅ (1+ i) ⇒ i R ⋅ (1+ i) log(1+ i) n
Example 5: Future value (p 433) Find the future value of an investment if $150 is deposited at the beginning of each month for 9 years and the interest rate is 7.2%, compounded monthly.
Example 6: Required payment (p 433) A company wants to have $450,000 after 2½ years to modernise its production equipment. How much of each previous quarter’s profits should be deposited at the beginning of the current quarter to reach this goal, if the company’s investment earns 6.8%, compounded quarterly? (hint: solve for R).
Present value of annuities (pp 437) How much do we have to deposit now in order to have so and so much later (An: total of money discounted at present)? To find that out we discount back. (1+ i)n −1 1 (compare with Table II appendix). An = R ⋅ n i ⋅(1+ i) Example 7: Present value (p 438) What is the present value of an annuity of $1,500 payable at the end of each 6-month period for 2 years if money is worth 8%, compounded semi-annually?
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Example 8: Payments from an annuity (p 439) A couple plans to set up an annuity with a $100,000 inheritance they received. What is the size of the quarterly payments they will receive for the next 6 years if the account pays 7%, compounded quarterly? (hint: solve for R).
Example 9: Number of payments from an annuity (p 440) An inheritance of $250,000 is invested at 9%, compounded monthly. If $2,500 is withdrawn at the end of each month, how long will it be until the account balance is $0? (hint: solve for n).
Example 10: Bond pricing (p 441) A 15-year corporate bond has a maturity value of $10.000 (S) and coupons at 5% (i) paid semi-annually. If an investor wants to earn a yield of 7.2% compounded semi-annually, what should he pay for this bond (P)?
Present value of annuities due: periodic payments, but at the beginning of each period. (1+ i)n −1 A(n,due) = R ⋅ ⋅ (1+ i) discounted at the present time with one extra payment. n i ⋅ (1+ i) Example 11: Lottery price (p 442) A lottery price worth $1,200,000 is awarded in payments of $10,000 at the beginning of each month for 10 years. Suppose money is worth 7.8%, compounded monthly. What is the real value of the price? (hint: solve for A(n,due)).
Deferred annuities: first payment not made at the first period, but k periods later for n periods. (1+ i)n −1 1 discounted at present time from period k. A(n,k ) = R ⋅ ⋅ n k i ⋅ (1+ i) (1+ i) Example 12: Lottery price payments (p 444) A lottery price of $500,000 is invested for future use. The winners plan to use the money for 8 semi-annual payments at the end of each 6month period after payments are deferred for 10 years. How much would each payment be if the money can be invested at 8.6% compounded semi-annually? (hint: solve for R).
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Solutions from in class examples (1+ 0.01)9 −1 1. S = $200 ⋅ = $1,873.71 (view Appendix) 0.01 2a)
(1 + 0.10) 8 − 1 = $22,871.78 0.10 S 36 = P(1 + i ) n = $22,871.78 ⋅ (1 + 0.10) 36 = $707,028 S = $2,000 ⋅
2b)
(1+ 0.10)36 −1 S = $2,000 ⋅ = $598,254 0.10
3.
$100,000 ⋅ 0.005 S ⋅i +1 log +1 log log (1.5) (1+ i) −1 S ⋅ i $1,000 R S = R⋅ ⇒ +1 = (1+ i)n ⇒ n = = = = 81.3 i R log(1+ i) log(1.005) log(1.005)
4.
n
It takes about 82 months, which is about 6.8 years. (1+ i)n −1 S ⋅i $300,000 ⋅ 0.03 S = R⋅ ⇒R= = = $26,169.15 semi-annual deposit. n i (1+ i) −1 (1+ 0.03)10 −1
5.
(1+ 0.006)108 −1 Sdue = $150 ⋅ ⋅ (1+ 0.006) = $22,836.59 0.006
6.
(1+ 0.017)10 −1 $450,000 ⋅ 0.017 $450,000 = R ⋅ = $40,967.39 ⋅ (1+ 0.017) ⇒ R = 10 0.017 (1+ 0.017) −1 ⋅ (1+ 0.017)
7.
1− (1+ i)−n 1− (1+ 0.04)−4 An = R ⋅ = $1,500 ⋅ = $5, 444.84 i 0.04
8.
1− (1+ i)−n An ⋅ i $100,000 ⋅ 0.0175 An = R ⋅ = = $5,138.57 ⇒ R = −n i 1− (1+ i) 1− (1+ 0.0175)−24
9.
10.
11.
R log 1− (1+ i)−n An ⋅ i 1 R R − An ⋅ i n An = R ⋅ = ⇒ = (1+ i) ⇒ n = ⇒ 1− i R (1+ i)n R − An ⋅ i log(1+ i) $2,500 log log(4) $2,500 − $250,000 ⋅ 0.0075 n= = = 185.53 ≈ 15.46 years log(1+ 0.0075) log(1.0075) S = P ⋅ (1+ i)n ⇒ $10,000 = P ⋅ (1+ 0.036)30 ⇒ P =
$10,000 = $3, 461.05 principal value (1+ 0.036)30
1− (1+ i)−n 1− (1+ 0.036)−30 An = R ⋅ = $250 ⋅ = $4,540.94 present value from coupons i 0.036 Total: $3,461.05 + $4,540.94 = 8,001.99 1− (1+ 0.0065)−120 A(120,due) = $10,000 ⋅ ⋅ (1+ 0.0065) = $836,843.55 0.0065
−n
12. A(n,k ) = R ⋅ 1− (1+ i) ⋅ (1+ i)−k ⇒ R =
i
A(n,k ) ⋅ i ⋅ (1+ i)k $500,000 ⋅ 0.043⋅ (1.043)20 = = $174,512.6 1− (1+ i)−n 1− (1.043)−8
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Applications in class (pp 434, pp 445) 6.3 7) Future value: Find the future value of an annuity of $1,300 paid at the end of each year for 5 years, if interest is earned at a rate of 6% compounded annually.
6.3 8) Future value: Find the future value of an annuity of $5,000 paid at the end of each year for 10 years, if interest is earned at a rate of 9% compounded annually.
6.3 10) Future value: Find the future value of an annuity of $300 paid quarterly for 5 years, if the interest rate is 12% compounded quarterly.
6.3 14) Time to reach a financial goal: If $4,000 is deposited at the end of each half year in an account that earns 6.2% compounded semi-annually, how long will it be before the account contains $120,000?
6.3 19) Annuities due: Find the future value of an annuity due of $100 each quarter for 2½ years at 12%, compounded quarterly.
6.3 23) Annuities due: How much must be deposited at the beginning of each year in an account that pays 8%, compounded annually, so that the account will contain $24,000 at the end of 5 years?
6.4 6) Present value: Find the present value of an annuity that pays $3,000 at the end of each 6-month period for 6 years with interest rate 6%, compounded semi-annually.
6.4 11) Retirement supplement: A retirement supplement contains $242,400. Suppose $200,000 is used to establish an annuity that earns 6%, compounded quarterly, and pays $4,500 at the end of each quarter. How long will it be until the account balance is contains $42,400?
6.4 21) Annuities due: What amount must be set aside now to have payments of $50,000 at the beginning of each year for the next 12 years if money is worth 5.92% compounded annually?
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Solutions 6.3 7)
(1+ 0.06)5 −1 S = $1,300 ⋅ = $7,328.22 0.06
6.3 8)
S = $5,000 ⋅
(1+ 0.09)10 −1 = $75,964.65 0.09
6.3 8)
S = $5,000 ⋅
(1+ 0.09)10 −1 = $75,964.65 0.09
6.3 10)
S = $300 ⋅
6.3 14) 6.3 19) 6.3 23)
(1+ 0.03)20 −1 = $8,061.11 0.03
120,000 ⋅ 0.031 log +1 4,000 log (1.93) n= = = 21.54 ≈ 10.77 years log(1+ 0.031) log(1.031) (1+ 0.03)10 −1 Sdue = $100 ⋅ ⋅ (1+ 0.03) = $1,180.78 0.03 (1+ i)n −1 Sdue $24,000 Sdue = R ⋅ = = $3, 787.92 ⋅ (1+ i) ⇒ R = n 5 i (1+ i) −1 (1+ 0.08) −1 ⋅ (1+ i) ⋅ (1+ 0.08) i 0.08
6.4 6)
1− (1+ 0.03)−12 An = $3,000 ⋅ = $29,862.01 0.03
6.4 11)
R $4,500 log log log(3) R − An ⋅ i $4,500 − $200,000 ⋅ 0.015 n= = = = 73.79 ≈ 74 quarters log(1+ i) log(1.015) log(1.015)
6.4 21)
1− (1.0592)−12 A(n,due) = $50,000 ⋅ ⋅ (1.0592) = $445,962.23 0.0592
Self-Evaluation exam II Time allowed: 20 Minutes Aids allowed: Non-programmable non-graphical calculator Use only the distributed pages
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8th Seminar Business Mathematics Loans and Amortisation Additional Material: Harshbarger, Reynolds: Chapter 6, Section 6.5. Odd numbered questions from exercises of section 6.5 (Solutions at pp A35 – A37)
Loans and Amortisation (pp 449) Amortisation: process of repaying a loan. 1− (1+ i)−n (1+ i)n −1 Present value of an ordinary annuity: An = R ⋅ (payment at end) = R ⋅ n i i ⋅ (1+ i) i ⋅ (1+ i)n i Size of payment: R = An ⋅ = A ⋅ n n 1− (1+ i)−n (1+ i) −1
An: Dept in $; R: Size of Payment;
i: Interest rate per period; n: Number of equal periodic payments Example 1: Payments (p 450) A dept of $1,000 with annual interest at 16%, compounded quarterly, is to be amortised by 20 quarterly payments (all the same size) over the next 5 years. What will the size of these payments be?
Example 2: Buying a Home (p 450) A man buys a house for $200,000. He makes a $50,000 payment and agrees to amortise the rest of the dept with quarterly payments over the next 10 years. If the annual interest on the dept is 12%, compounded quarterly, find …. a) … the size of the quarterly payments. b) … the total amount of payments. c) … the total amount of extra payments on top of $200,000.
Example 3: Affordable Home (pp 450) Chuckie and Angelica have $30,000 for down payment, and their budget can accommodate a monthly mortgage payment of $1,200. What is the most expensive home they can buy if they can borrow money for 30 years at 7.8% annual interest, compounded monthly?
Amortisation schedule: summarises the information regarding the amortisation of a loan, which is: Period, Payment, Interest, Balance Reduction, and Unpaid Balance.
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Example 4: Amortisation Schedule (p 451) Loan: $10,000; Annual Interest: 10%; n = 5 equal annual payments.
Unpaid Balance of a Loan (payoff amount, outstanding principal of the loan): present value needed to generate all the remaining payments. With other words: the amount we still have to pay after a certain period of time we paid in the past.
⎡1 − (1 + i) −( n−k ) ⎤ ⎡ (1 + i) ( n−k ) − 1⎤ Unpaid Balance An−k = R ⋅ ⎢ . = R ⋅ ⎥ ⎢ ( n−k ) ⎥ i ⎣ ⎦ ⎣ i ⋅ (1 + i) ⎦ Example 5: Unpaid Balance (p 453) Continues from Example 2 (A man buys a house for $200,000. He makes a $50,000 payment and agrees to amortise the rest of the dept with quarterly payments over the next 10 years. If the annual interest on the dept is 12%, compounded quarterly.) The payment for a loan of $150,000 at 12%, compounded quarterly, for 10 years is $6489.36 per quarter. Find the unpaid balance immediately after the 15th payment.
Example 6: Effect of Paying an Extra Amount (p 453) An amortisation schedule for the first 24 monthly payments on a $100,000 loan amortised for 30 years at 6% annual interest, compounded monthly is shown in the table below (for the full table compare with “AmortisationSchedule.xlsx” Example 6). The unpaid balance after 24 monthly payments is $97,468.25. Suppose that from this point the borrower decides to pay $650 per month. 6a) How many more payments must be made? 6b) How much would this save over the life of the loan? Payment Number 0 1 2 ... 24
Payment Balance Unpaid Total Amount Interest Reduction Balance Interest $0 $0 $0 $100000 $0.00 $599.55 $500.00 $99.55 $99900.45 $500.00 $599.55 $499.50 $100.05 $99800.40 $999.50 ... ... ... ... ... $599.55 $487.90 $111.65 $97468.25 $11857.45
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Solutions from in class examples i ⋅ (1+ i)n 0.04 ⋅ (1.04)20 1) R = An ⋅ = $1,000 ⋅ = $1,000 ⋅ 0.0736 = $73.58 n 20 (1+ i) −1 (1.04) −1 2a) $50,000 paid at the beginning, n = 4 payments per year for 10 years = 40, i = 0.12/4 = 0.03 0.03⋅ (1.03)40 R = $150,000 ⋅ = $150,000 ⋅ 0.0433 = $6489.36 40 (1.03) −1 2b) Total payments = 40 ⋅ $6489.36 + $50,000 = $259,574.40 + $50,000 = $309,574.4 2c) Remaining interest payments = $309,574.4 − $200,000 = $109,574.4 3) $30,000 paid at the beginning; R = $1,200; n = 12 payments for 30 years = 360; i = 0.078 / 12 = 0.0065 (1+ i)n −1 (1.0065)360 −1 An = R ⋅ $1,200 ⋅ = $1,200 ⋅138.914 = $166,696.65 n 360 i ⋅ (1+ i) 0.0065⋅ (1.0065) Total amount = $166,696.65 + $30,000 = $196,696.65 i ⋅ (1+ i)n 0.10 ⋅ (1.10)5 4) R = An ⋅ = $10,000 ⋅ = $10,000 ⋅ 0.263797 = $2,637.97 n 5 (1+ i) −1 (1.10) −1 Period Payment Interest Balance Reduction Unpaid Balance 1 $2,637.97 $1,000 $1,637.97 $8,362.03 2 $2,637.97 $836.20 $1,801.77 $6,560.26 3 $2,637.97 $656.03 $1,981.94 $4,578.32 4 $2,637.97 $457.83 $2,180.14 $2,398.18 5 $2,638.00 $239.82 $2,398.18 $0.00 Total $13,189.88 $3,189.88 $10,000.00 Last payment increased by 3 cent to get a balance of $0.
5) Unpaid balance = present value of annuity with 40 – 15 = 25 payments remaining. R = $6,489.36; n – k = 25; i = 0.03. ⎡1 − (1 + i) − ( n−k ) ⎤ ⎡1 − (1.03) −25 ⎤ An−k = R ⋅ ⎢ = $ 6 , 489 . 36 ⋅ ⎥ ⎢ ⎥ = $6,489.36 ⋅17.413 = $113,000.18 i ⎣ ⎦ ⎣ 0.03 ⎦ 6a) Solve “An” formula for n. An = $97,468.25; R = $650; i = 0.005. (1+ i)n −1 R An = R ⋅ ⇒ An ⋅ i ⋅ (1+ i)n = R ⋅ (1+ i)n −1 ⇒ (1+ i)n ⋅ (R − An ⋅ i) = R ⇒ (1+ i)n = ⇒ n ⋅ log(1+ i) = n R − An ⋅ i i ⋅ (1+ i) R $650 $650 log log log R R − An ⋅ i $650 − $97, 468.25⋅ 0.005 $162.6588 0.6016 log = = = ≈ 277.8 ⇒ n= log(1+ i) log(1.005) log(1.005) 0.002 R − An ⋅ i
6b) To determine the savings, the two payment schedules are compared as: b1) The payment consists of n = 277.8 payments of $650 and the first 24 of $599.55. b2) The original n = 360 payments consist of $599.55.
b1) Total Payment = 277.8⋅ $650 + 24 ⋅ 599.55 = $194,959.20 b2) Total Payment = 360 ⋅ 599.55 = $215,838.00 Savings = $215,838.00 − $194,959.20 = $20,878.80
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Applications in class (pp 454) 6.5 4) Amortise a Loan: A loan of $10,000 is to be amortised with 10 equal quarterly payments. If the annual interest rate is 6%, compounded quarterly, what is the periodic payment?
6.5 8) Borrow: Adrianne’s Auto Repair wants to add a new service bay. How much can they borrow at 5% annual interest, compounded quarterly for 4.5 years, if the desired quarterly payment is $6000?
6.5 26) Unpaid Balance: A company that purchases a piece of equipment by borrowing $250,000 for 10 years at 6% annual interest, compounded monthly, has monthly payments of $2,775.51. a) Find the unpaid balance on this loan after 1 year. b) During the first year, how much interest does the company pay?
6.5 36) Home Mortgage: Clark and Lana take a 30-year home mortgage of $121,000 at 7.8% annual interest, compounded monthly. They make their regular monthly payments for 5 years, then decide to pay $1,000 per month. a) Find their regular monthly payment. b) Find the unpaid balance when they begin paying the $1,000. c) How many payments of $1,000 will it take to pay off the loan? d) How much interest will they save by paying the loan this way? Solutions from Applications in class i ⋅ (1+ i)n 0.015⋅ (1.015)10 6.5 4) R = An ⋅ = $10,000 ⋅ = $10,000 ⋅ 0.10843 = $1,084.34 n 10 (1+ i) −1 (1.015) −1 6.5 8) R = $6,000; n = 4 payments for 4.5 years = 18; i = 0.05 / 4 = 0.0125 (1+ i)n −1 (1.0125)18 −1 An = R ⋅ = $6,000 ⋅ = $6,000 ⋅16.030 = $96,177.29 n 18 i ⋅ (1+ i) 0.0125⋅ (1.0125) 6.5 26) Compare with “AmortisationSchedule.xlsx” Example 6.5.26) a) $231,181.98; b) $14,488.10 6.5 36) Compare with “AmortisationSchedule.xlsx” Example 6.5.36) a) $ 943.09; b) $112212.10; c) ⎛ ⎞ R $1000 ⎞ ⎛ $1000 ⎞ ⎟⎟ log⎛⎜ log⎜⎜ ⎟ log⎜ ⎟ R − An ⋅ i ⎠ $943.09 − $112,212.10 ⋅ 0.0065 ⎠ $213.71135 ⎠ 0.6702 ⎝ ⎝ ⎝ n= = = = ≈ 238.18.41 log(1 + i) log(1.0065) log(1.0065) 0.0028 b1) Total Payment = 229.13⋅ $1000 + 60 ⋅ 943.09 = $194,959.20 d) b2) Total Payment = 360 ⋅ 599.55 = $215,838.00
Savings = $215,838.00 − $194,959.20 = $20,878.80
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9th Seminar Business Mathematics Review of Course Business Mathematics Applications Business Mathematics: Compare with “Questions for Course Repetition”.
Review of Course Business Mathematics Theory Repetition Foundation Maths n a n an n−m n n n a a • a = a ; m = a ; a • b = (ab) ; n = ; (a m ) n = a m•n a b b n n n bc = b ⋅ c if b, c ∈ ℜ n
€
n
m n m
m
n +m
b nb = c nc
if b, c ∈ ℜ and c ≠ 0
b m−n = b b
if b ∈ ℜ and b ≠ 0
b ⋅ n b = m+n b
m n
b = m⋅n b Composite functions g of f (g f ) and f of g ( f g) → (g f )(x) = g( f (x)) and ( f g)(x) = f (g(x)) b x = 0 ⇒ (0;b); y = 0 ⇒ − ;0 a . Linear function: f (x) = mx + b ; intercepts: Breakeven Analysis (pp 119) C(x) = Total cost (fixed cost + variable cost) = cfix + cvar R(x) = Total revenue from sale of x units = (price per unit) • (number of units) P(x) = Profit from sale of x units P(x) = R(x) – C(x) Break-even point: cost = revenue:
R(x) = C(x)
Marginal profit (MP) = profit of one additional unit (slope of profit function) Marginal cost (MC) = cost to produce one additional unit (slope of cost function) Marginal revenue (MR) = revenue of selling one additional unit (slope of revenue function) Market Equilibrium Analysis (pp 122) At market equilibrium demand = supply (market cleared). Law of demand: as demand increases the price increases as well. Law of supply: as supply increases, the price decreases as well. Quantity demanded or supplied (q): independent variable; price (p): dependent variable.
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At market equilibrium, the equilibrium quantity and the equilibrium price apply. Demand and supply curves intersect. Parabolas (pp 151) General form: f (x) = ax 2 + bx + c
b b Vertex point for f (x) = ax 2 + bx + c → ... → − ; f − setting y = 0 gives two null 2a 2a (intercepts). Because of symmetry the vertex has to be in the middle of these two points. f (b) − f (a) . Average rate of change = b−a Exponential functions (pp 357)
f (x) = a x with a ≠ 0 ; f (x) = ca x ; f (x) = ca − x ; f (x) = ca bx Exponential growth function: f (x) = Ca x . Logarithmic functions (pp 369) for y = log a (x) with a > 0 and a ≠ 1 (logarithmic form) ⇒ x = a y (exponential form) log(x) ln(x) Change of base of logarithmic functions: log b (x) = = log(b) ln(b)
f (x) = ln(x) ↔ f (x) = e x
Some inverse functions and their original functions: f (x) = log(x) ↔ f (x) = 10 x f (x) = sin(x) ↔ f (x) = arcsin(x) Theory Repetition Finance Maths Simple Interest; Sequences (pp 404) Simple Interest I from a sum of money P (principal) is given by: I = P ⋅ r ⋅ t r : annual interest rate (as decimal); t: time (in years) Simple interest just “adds” the interest rate to the principal money each year. It disregards the “interest of the interest” rate. Future value of investment or loan: S = P + I
Arithmetic Sequences (pp 407) Common differences: an = an−1 + d for n > 1 ; an = a1 + (n −1)⋅ d ; Sn =
n ⋅ (a1 + an ) 2
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Compound Interest (pp 412) Compound interest adds the interest of each period to the principal. mt
r r i = ; n = mt; S = P ⋅ (1+ i)n = P ⋅ 1+ m m Continuous compounding: S = P ⋅ ert Euler's number: e = 2.7182818... Annual Percentage Yield (APY) Interest compounded: m r Periodical: APY = 1+ −1 = (1+ i)m −1 , Simple: APY = mi , Continuous: APY = er −1 . m Geometric Sequences a1 ⋅ (1− r n ) n−1 Common ratio: an = r ⋅ an−1 for n > 0 . nth term: an = a1 ⋅ r ; Sn = with r ≠ 1 . 1− r Future value of annuities (pp 427) Annuities: an annual amount of money deposited with a certain interest rate for later use. When you deposit the same amount of money regularly at the same interest, this follows the law of a (1+ i)n −1 geometric sequence. S = R ⋅ ; S: future value of annuities; R: regularly deposited i (1+ i)n −1 amount; i: interest rate. s¬ = = future value, if $1 is deposited. ni i Annuities due: periodic payments are made at the beginning of each period, hence, interest for (1+ i)n −1 one more period Sdue = R ⋅ ⋅ (1+ i) . Solve for n, if the time is asked: i S ⋅i log due +1 n (1+ i) −1 Sdue ⋅ i R ⋅ (1+ i) Sdue = R ⋅ +1 = (1+ i)n ⇒ n = ⋅ (1+ i) ⇒ i R ⋅ (1+ i) log(1+ i) Present value of annuities (pp 437) How much do we have to deposit now in order to have so and so much later? To find that out we discount back. For such a geometric sequence, the term is interesting, not the sum. 1− (1+ i)−n (1+ i)n −1 (compare with Table II appendix). An = R ⋅ = R ⋅ n i i ⋅ (1+ i) Present value of annuities due: periodic payments, but at the beginning of each period. 1− (1+ i)−n (1+ i)n −1 A(n,due) = R ⋅ ⋅ (1+ i) = R ⋅ ⋅ (1+ i) n i i ⋅ (1+ i) Deferred annuities: first payment not made at the first period, but k periods later. 1− (1+ i)−n (1+ i)n −1 1 −k A(n,k ) = R ⋅ ⋅ ⋅ (1+ i) = R ⋅ n k i i ⋅ (1+ i) (1+ i)
