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2
Introduction to
Mathematical
Economics
OVERVIEW
Managerial economics is the synthesis of microeconomic theory, mathematics and statistical methods to find optimal solutions to managerial
decision-making problems. As an applied discipline, managerial economics
integrates economic theory with the techniques of quantitative analysis,
including mathematical economics, optimization analysis, regression analysis, forecasting, linear programming, and risk analysis. Managerial economics attempts to demonstrate how the optimality conditions postulated in
economic theory can be applied to “real world” business situations to optimize firms’ organizational objectives. This chapter reviews the fundamental mathematical concepts encountered throughout the text.
Every student brings something of himself or herself to the study of managerial economics. In many engineering programs, for example, students are
required to take finance and economics courses to develop an understanding of the business aspects of research, development, construction and
product development. In general, these students come with rich backgrounds in quantitative methods, and perhaps somewhat deficient in economic principles. For students who fall into this category, a very brief review
of the topics presented in this chapter may be all that is necessary before
moving to chapters that are more specifically economic in nature. For many
liberal-arts-turned-business majors, however, a more thorough examination
of mathematical methods may be absolutely essential to successfully master
the material presented in subsequent chapters of this text.
Managerial Economics: Theory and Practice
15
Copyright © 2003 by Academic Press.
All rights of reproduction in any form reserved.
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Introduction to Mathematical Economics
MULTIPLE CHOICE QUESTIONS
2.1
Consider the relationship y = f(x). In this relationship:
A. y is called the dependent variable and x is called the
independent variable.
B. y is called the independent variable and x is called the
dependent variable.
C. y is called the range and x is called the domain.
D. y is called the domain and x is called the range.
2.2
Consider the relationship y = f(x).
A. The notation f can be regarded as a specific rule that defines the
relationship between values of x and values of y.
B. The notation f is a rule that maps values of x into a value of y.
C. This relationship may also be expressed as f: x Æ y.
D. All of the above are correct.
2.3
Consider the functional relationship y = f(x) = 10 + 5x - 2.5x2. At
x = 10, the value of y is:
A. -190.
B. -290.
C. 290.
D. 310.
2.4
Suppose that the demand for a product is given by the functional
relationship Q = 500 - 4P. If a firm wants to sell 100 units, the what
price should it charge?
A. $10
B. $50
C. $100
D. $150
2.5
Consider the functional relationship y = f(x) = a + bx. The value b is
called a:
A. Tangent.
B. Intercept.
C. Slope.
D. Parameter.
E. Both C and D are correct.
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Multiple Choice Questions
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2.6
Suppose that we know that the quantity demanded of a good is a
linear function of its selling price. Suppose that (Q1, P1) = (95, 1) and
(Q2, P2) = (75, 5). The demand equation is:
A. Q = 100 - 5P
B. Q = 200 - 4P
C. Q = -100 + 2P
D. Q = 50 + 2.5P
2.7
Consider the functional relationship y = f(x) = a + bx. The value b is:
A. Dy/Dx
B. (y2 - y1)/(x2 - x1)
C. [f(x2) - f(x1)]/(x2 - x1)
D. All of the above are correct.
2.8
The market demand and supply equations for a product are given
by the equations QD = 20 - 2P and QS = -10 + 3P, respectively. The
equilibrium price and quantity are:
A. P* = 4 and Q* = 6.
B. P* = 5 and Q* = 7.
C. P* = 6 and Q* = 8.
D. P* = 7 and Q* = 9.
2.9
Suppose that the relationship between total profit (p) and output
(Q) is given by the expression p = f(Q). In the expression, Q is
described as the:
A. Dependent variable.
B. Independent variable.
C. Slope.
D. Intercept.
E. Parameter.
2.10 Consider the relationship y = f(x). The inverse of this function exists
when:
A. A one-to-one correspondence exists between x and y.
B. The function is both one-to-one and onto.
C. The function has a first derivative.
D. The function has a second derivative.
E. Both A and B are correct.
2.11 A monotonically increasing function:
A. Has an inverse.
B. Is a one-to-one correspondence.
C. Is both one-to-one and onto.
D. All of the above are correct.
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Introduction to Mathematical Economics
2.12 The value x-4 may also be written as:
A. 1/x4
B. 4÷x
C. -(x)1/4
D. e-4x
2.13 The function y = a + bx + cx2:
A. Is a linear equation.
B. Is a quadratic equation.
C. Is a cubic equation.
D. Has a graph called a hyperbola.
E. Both B and D are correct.
2.14 Optimization analysis involves:
A. Maximizing some objective function.
B. Minimizing some objective function.
C. Maximizing or minimizing some objective function.
D. Determining the first derivative of a quadratic function.
2.15 The derivative of a function:
A. Is the slope of the function when the interval between adjacent
values of the independent variable is infinitesimally small.
B. Is constant for all values of the independent variable for
nonlinear functions.
C. Exists for all values of the independent variable for
discontinuous functions.
D. Only exists for inverse functions.
E. None of the above.
2.16 The first derivative of the function y = f(x) = 0.5x40 - x22 + 5x2 + 5 is:
A. dy/dx = f¢(x) = 0.5x41 - 22x23 + 10x
B. dy/dx = f¢(x) = 20x41 - 22x23 + 10x
C. dy/dx = f¢(x) = 20x39 - 22x21 + 10x
D. dy/dx = f¢(x) = 20x39 - 22x21 + 10x
2.17 The first derivative of the function y = f(x) = 2x2/3x3 is:
A. dy/dx = f¢(x) = (-6x4)(9x6)-1
B. dy/dx = f¢(x) = -36x24
C. dy/dx = f¢(x) = 36x24/9x6
D. dy/dx = f¢(x) = 1
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Multiple Choice Questions
19
2.18 The first derivative of the function y = e4x is:
A. dy/dx = f¢(x) = e4x.
B. dy/dx = f¢(x) = 4e4x.
C. dy/dx = f¢(x) = 4ex.
D. dy/dx = f¢(x) = 4ex.
2.19 The first derivative of y = 3x/(3x + x2) is:
A. [(3 + 2x)3x + 3[3x + x2)]/(3x + x2)2.
B. (3 + 2x)3x/(3x + x2)2.
C. -3/(3 + x)2.
D. [(3 + 2x)3x/(3x + x2)2.
2.20 The function y = f(x) = x + 5x9 + x13:
A. Is monotonically increasing.
B. Is monotonically decreasing.
C. Is monotonically increasing for positive values of x and
monotonically decreasing for negative values of x.
D. Is neither monotonically increasing nor decreasing.
2.21 Consider the function y = f(x) = x + 5x9 + x13. The derivative dx/dy is:
A. 1 + 45x8 + 13x12.
B. (1 + 45x8 + 13x12)-1.
C. -(1 + 45x8 + 13x12)-2.
D. None of the above.
2.22 When the slope of the average cost function is positive, then:
A. The slope of the marginal cost function must also be positive.
B. The slope of the marginal cost function must also be negative.
C. Marginal cost must be greater than average cost.
D. Marginal cost must be less than average cost.
2.23 Average profit is maximized when:
A. Marginal profit is falling.
B. Marginal profit is maximized.
C. Marginal profit is rising.
D. Marginal profit is equal to average profit.
2.24 When average total cost is rising, then:
A. Marginal cost must be rising;
B. Marginal cost must be greater than average total cost.
C. Marginal cost must be less than average total cost.
D. Marginal cost must be falling.
E. Marginal cost must be maximized.
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Introduction to Mathematical Economics
2.25 The function y = f(x) = 10 - 2x has a maximum value when x equals:
A. 0.
B. 5.
C. 10.
D. This function is linear. It has neither a maximum nor a minimum
value.
2.26 The function y = f(x) = 100 - 25x + 2.5x2 has a maximum value when
x equals:
A. 5.
B. 25.
C. 50.
D. This function does not have a maximum value.
2.27 Suppose that the total revenue function of a firm is given by the
expression TR = 500Q - 5Q2. The value for Q which total revenue is
optimized is:
A. 10.
B. 50.
C. 100.
D. This function does not have a maximum value.
2.28 For a nonlinear functions to have a local maximum, then:
A. The first derivative is zero and the second derivative is zero.
B. The first derivative is zero and the second derivative is positive.
C. The first derivative is zero and the second derivative is negative.
D. The first derivative is positive and the second derivative is
positive.
E. The first derivative is positive and the second derivative is
negative.
2.29 Suppose that a firm’s total profit function is p = 100x + 68y - 2xy 5x2 - 5y2. The profit maximizing combination of x and y is:
A. x = 9 and y = 12.
B. x = 9 and y = 5.
C. x = 7 and y = 9.
D. x = 12 and y = 9.
2.30 Suppose that a firm’s total profit function is p = 100x + 68y - 2xy 5x2 - 5y2. The firm’s maximum profit is:
A. $620.
B. $780.
C. $940.
D. $1,000.
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Multiple Choice Questions
21
2.31 A firm’s total profit function is p = x2 + y2 - xy, where x and y
represent two goods produced on the same assembly line. If
management is required to produced a total number of 10 units,
then the profit-maximizing combination of goods x and y are:
A. x = 7 and y = 3.
B. x = 6 and y = 4.
C. x = 5 and y = 5.
D. x = 4 and y = 6.
E. x = 3 and y = 7.
2.32 You must produce 500 units of goods x and y this month. Using the
Lagrangian multiplier method, to determine that the profit
maximizing combination of the two goods. You also learn that the
value of the Lagrangian multiplier is $5.00. From this you conclude
that:
A. Marginal revenue is $5.00.
B. Average revenue is $5.00.
C. Increasing the number total production of x and y produced by
one unit will increase profits by $5.00.
D. Average profit is $5.00.
2.33 Using the Lagrangian multiplier method, to determine the optimal
allocation of your $5,000,000 advertising budget between magazines,
radio and television. You also learn that the value of the Lagrangian
multiplier is -$100. From this you conclude that:
A. Increasing your advertising budget by one dollar will lower
profits by $100.
B. Decreasing your advertising budget by one dollar will increase
profits by $100.
C. Advertising has a negative effect on profits.
D. Both A and B are correct.
2.34 The indefinite integral of the expression f¢(x) = 6x2 + 0.5x + 50 is:
A. 2x3 + 0.25x2 + 50x + c.
B. 2x3 + 0.25x2 + 50x.
C. 2x3 + 0.25x2 + c.
D. 2x3 + 0.25x2.
2.35 An integral may be interpreted as:
A. The inverse of the objective function.
B. The Lagrangian multiplier.
C. The area under a curve.
D. A monotonic function.
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Introduction to Mathematical Economics
SHORTER PROBLEMS
2.1
Find the first and second derivatives of the following functional
relationships:
A. y = f(x) = -25 + 3x12 - 5x3
B. y = f(x) = 4ex - 2x5 + 17x23
C. y = f(x) = 17x-7 + ÷(x) - (3x -5)3
D. y = f(x) = loge(4x5 + 2x)
2.2
The market demand and supply equations for a product are
QD = 350 - 5P
QS = -100 + 2.5P
where Q is quantity and P is price. Determine the equilibrium price
and quantity.
2.3
In macroeconomic theory, the spending multiplier (k) is given by the
expression S = 1/(1 - MPC), where the MPC is the marginal
propensity to consume.
A. Suppose that MPC = 0.9. Calculate the value of the spending
multiplier (S).
B. Demonstrate that the spending multiplier is the sum of an
infinite geometric progression.
2.4
Consider the function
y = f(x) = 2x + 0.5x5 + 3x9
A. Is this function monotonic?
B. If the function is monotonic, then use the inverse-function rule
to find dx/dy.
2.5
Consider the function
y = f(x) = -(x + x6)
A. Is this function monotonic?
B. If the function is monotonic, then use the inverse-function rule
to find dx/dy.
2.6
The total revenue and total cost functions of a firm are
TR(Q) = PQ = 150Q - 5Q2
TC(Q) = 2.5Q2
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Longer Problems
A. Determine the profit (p) maximizing output level where
p(Q) = TR(Q) - TC(Q).
B. Determine maximum p.
C. Determine the price per unit at which the p maximizing output
is sold.
2.7
The demand equation for a firm’s output is
Q = f(P, A) = 80P - 2P2 - PA - 3A2 + 100A
where Q represents units of output, P the selling price of the
product, and A represents advertising expenditures ($000s).
A. Determine the price and advertising expenditures necessary to
will maximize the firm’s sales.
B. Determine maximum Q.
2.8
Consider the following objective function:
y = 150 + 30x - 5x2
A. Find the value of x that results in an extreme value of y.
B. Is the value of y a minimum or a maximum?
C. What is the minimum or maximum value of y?
2.9
Evaluate the following indefinite integrals:
A. Ú(10x3 + 600x2 - 50)dx
B. Ú(6x2 + 3x)dx
C. Ú(5x - 25)dx
LONGER PROBLEMS
2.1
The total revenue (TR) and total cost (TC) functions of a
monopolist are:
TR(Q) = PQ = 50Q - 2.5Q2
TC(Q) = 100 + 25Q - 5Q2 + 0.5Q3
A. Determine the profit (p) maximizing output level where
p(Q) = TR(Q) - TC(Q).
B. Determine maximum p.
C. Determine the price per unit at which the p maximizing output
is sold.
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2.2
Introduction to Mathematical Economics
Suppose that the demand for a firm’s product is
Q = f(P, T, M) = -2P + P2 - TM + 5T + 6M - T2 - M2
where Q represents unit sales, P represents the price of the product,
T ($000s) represents television advertising expenditures and M
($000s) represents magazine advertising expenditures.
A. Determine P, T, and M that maximizes unit sales.
B. Calculate the maximum unit sales.
2.3
Consider a profit-maximizing firm that produces two products, x and
y, on the same assembly line. The firms profit function is given by
the equation
p(x, y) = 200x - 5xy + 100y
A. Using the substitution method, determine profit maximizing
output levels of x and y subject to the production requirement
that total output equal 50 units.
B. Calculate the firm’s profits.
2.4
Consider a profit-maximizing firm that produces two products, x and
y, on the same assembly line. The firms profit function is given by
the equation
p(x, y) = 200x - 5xy + 100y
A. Using the Lagrange multiplier method, determine profit
maximizing output levels of commodities x and y subject to the
production requirement that total output equals 50 units. What is
the value of the Lagrange multiplier?
B. Calculate the firm’s profits.
C. Interpret the Lagrange multiplier.
D. What would be the firm’s profits be if there was no production
requirement?
ANSWERS TO MULTIPLE CHOICE QUESTIONS
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
B.
D.
A.
C.
E.
A.
D.
C.
2.9
2.10
2.11
2.12
2.13
2.14
2.15
B.
E.
D.
A.
B.
C.
A.
2.16
2.17
2.18
2.19
2.20
2.21
2.22
D.
A.
B.
C.
A.
B.
C.
2.23
2.24
2.25
2.26
2.27
2.28
2.29
D.
B.
D.
D.
B.
C.
B.
2.30
2.31
2.32
2.33
2.34
2.35
2.35
A.
C.
C.
D.
A.
C.
C.
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Solutions to Shorter Problems
25
SOLUTIONS TO SHORTER PROBLEMS
2.1
A. dy/dx = f¢ (x) = 36x11 - 15x2
d2y/dx2 = f≤ (x) = 396x10 - 30x
B. dy/dx = f¢ (x) = 4ex - 10x4 + 391x22
d2y/dx2 = f≤ (x) = 4ex - 40x3 + 8602x21
C. dy/dx = f¢ (x) = -119x-8 + 0.5/÷x - 9(3x - 5)2
d2y/dx2 = f≤ (x) = 952x-9 - 0.25x-1.5 - 164x + 270
D. dy/dx = f¢ (x) = (20x4 + 2)(4x5 + 2x)-1
d2y/dx2 = f≤ (x) = -(20x4 + 2)2(4x5 + 2x)-2 + 80x3(4x5 + 2x)-1
2.2
The equilibrium price and quantity are given as
QD = QS
Substituting into this expression we have
350 - 5P = -100 + 2.5P
P* = (350 + 100)/7.5 = 60
Q* = 350 - 5(60) = -100 + 2.5(60) = 50
2.3
A. S = 1/(1 - MPC) = 1/(1 - 0.9) = 1/0.1 = 10
B. The sum of the infinite geometric progression is
R = a + ar + ar2 + ar3 . . . = a(1 + r + r2 + r3 . . .)
When r > 1, then R = •. Assume that 0 < r < 1. Define
(1) S = 1 + r + r2 + r3 . . .
Multiplying this expression by r we obtain
(2) rS = r + r2 + r3 + r4 . . .
Subtracting equation (2) from equation (1) we obtain
S - rS = (1 + r + r2 + r3 . . .) - (r + r2 + r3 + r4 . . .)
= 1 + r + r2 + r3 . . . -r - r2 - r3 - r4 . . . = 1
Solving for S we obtain
S = 1/(1 - r) = 1/(1 - 0.9) = 10
2.4
A. The derivative of this function is
dy/dx = f¢(x) = 2 + 2.5x4 + 27x8
which is positive for all values of x. Thus, the function f(x) is a
monotonically increasing function.
B. Because f(x) is a monotonically increasing function then the
inverse function g(y) = f -1(y) exists. Thus, it is possible to use the
inverse-function rule to determine the derivative of the inverse
function, i.e.,
dx/dy = 1/(dy/dx) = 1/(2 + 2.5x4 + 27x8)
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Introduction to Mathematical Economics
2.5
A. The derivative of this function is
dy/dx = f ¢(x) = -(1 + 6x5 )
This function is not monotonic since the sign of dy/dx depends
on whether x is positive or negative. On the other hand, the
derivative is negative for all positive values for x.
B. Because the derivative of f(x) is negative for all x > 0, then it is
possible to use the inverse-function rule to determine the
derivative of the inverse function, i.e.,
g¢(y) = dx/dy = 1/(dy/dx) = 1/f ¢(x) = 1/ - (1 + 6x5 )
for all x > 0.
2.6
A. p = 150Q - 5Q2 - 2.5Q2 = 150Q - 7.5Q2
dp/dQ = 150 - 15Q = 0, i.e., the first-order condition for a profit
maximum.
Q* = 10 units
d2p/dQ2 = -15 < 0, i.e., the second-order condition for a profit
maximum.
B. B = 150(10) - 7.5(10)2 = $750
C. TR = PQ = 150Q - 5Q2 = (150 - 5Q)Q
P = 150 - 5Q = 150 - 5(10) = $100
2.7
A. The first-order conditions for sales maximization are
∂Q/∂P = 80 - 4P - A = 0
∂Q/∂A = -P - 6A + 100 = 0
Assume that the second-order conditions for sales maximization
are satisfied. Solving these equations simultaneously we obtain
P* = $15.82
A* = $13.33 (000’s)
B. Q* = 80(15.82) - 2(15.82)2 - (15.82)(13.33) - 3(13.33)2
+ 100(13.33)
= 1,265.6 - 500.54 - 210.88 - 533.07 + 1,333 = 1,354.11
2.8
A. The first-order condition is for an extreme value of y is
dy/dx = f¢ (x) = 30 - 10x = 0
Solving for x we obtain
x* = 3
B. The second-order condition is
d2y/dx2 = f≤ (x) = -10 < 0
Since the second derivative of the objective function is negative,
the value of y reaches a maximum value at x* = 3.
C. y* = 50 + 30(3) - 5(3)2 = 195
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Solutions to Longer Problems
2.9
27
A. y = (10/4)x3+1 + (600/3)x2+1 - (50/1)x0+1 + c = 2.5x4 + 200x3
- 50x + c
B. y = (6/3)x2+1 + (3/2)x1+1 + c = 2x3 + 1.5x2 + c
C. y = (5/2)x1+1 - (25/1)x0+1 + c = 2.5x2 - 25x + c
In each of the above cases, c is an unknown constant. The above
results may be verified by taking the first derivative of each
equation to recover the original expression to be integrated.
SOLUTIONS TO LONGER PROBLEMS
2.1
A. p = TR - TC = 150Q - 2.5Q2 - 100 - 25Q + 5Q2 - 0.5Q3
= -100 + 125Q + 2.5Q2 - 0.5Q3
dp/dQ = 125 + 5Q - 1.5Q2 = 0, i.e., the first-order condition for a
local maximum
The quadratic formula is given by the expression
Q1,2 = {-b ± ÷[b2 - 4ac]}/2a, where a = -1.5, b = 5 and c = 125.
Substituting, we obtain
Q1,2 = {-5 ± ÷[(5)2 - 4(-1.5)(125)]}/2(-1.5)
= [-5 ± ÷(25 + 750)]/2(-1.5)
= (-5 ± ÷775)/-6
= (-5 ± 27.84)/-6
Q1 = (-5 - 27.84)/-6 = -32.84/-6 = 5.47
Q2 = (-5 + 27.84)/-6 = -6/-6 = -3.81
To determine whether these values constitute a minimum or a
maximum, we can substitute these values into the profit function
and determine the minimum and maximum values directly, or we
can examine the values of the second derivatives:
d2p/dQ2 = 5 - 3Q
For Q1 = 5.47,
d2p/dQ2 = 5 - 3(5.47) = -11.41 < 0, i.e., the second-order
condition for a local maximum.
For Q2 = -3.81,
d2p/dQ2 = 5 - 3(-3.81) = 16.43 > 0, i.e., the second-order
condition for a local minimum.
Substituting Q1 = 5.47 into the p function yields a maximum
profit of
B. p* = -100 + 125(5.47) + 2.5(5.47)2 - 0.5(5.47)3 = $576.72
C. TR(Q) = PQ = 150Q - 2.5Q2 = (150 - 2.5Q)Q
P* = 150 - 2.5(5.47) = $136.33
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Introduction to Mathematical Economics
2.2
A. The first-order conditions for unit sales maximization are:
∂Q/∂P = -2 + 2P = 0
∂Q/∂T = -M + 5 - 2T = 0
∂Q/∂M = -T + 6 - 4M = 0
Solving the first-order conditions simultaneously, the solution
values after rounding are:
P* = $1
T* = $2 (000s)
M* = $1 (000s)
B. Q* = -2(1) + (1)2 - (2)(1) + 5(2) + 6(1) - (2)2 - (1)2 = 8 (000)
units
2.3
A. This is a constrained maximization problem that may be
formally expressed as
Maximize: p(x, y) = 200x - 5xy + 100y
Subject to: x + y = 50
Solving for x and substituting into the objective function we
obtain
p(y) = 200(50 - y) - 5(50 - y)y + 100y
= 10,000 - 350y + 5y2
The first-order condition is
dp/dy = - 350 + 10y = 0
y* = 35
x* = 50 - y = 50 - 35 = 15
The second-order condition for a local maximum is satisfied
since
d2p/dy2 = 10 > 0
B. p* = 200(15) - 5(15)(35) + 100(35) = $3,875
2.4
A. This is a constrained maximization problem that may be
formally expressed as
Maximize: p(x, y) = 200x - 5xy + 100y
Subject to: x + y = 50
The Lagrange expression is:
(x, y) = 200x - 5xy + 100y + l(50 - x - y)
The first-order conditions are:
∂/∂x = x = 200 - 5y - l = 0
∂/∂y = y = -5x + 100 - l = 0
∂/∂l = l = 50 - x - y = 0
This system of three linear equations in three unknowns can be
solved for the following values:
x* = 15
y* = 35
l* = $25
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Solutions to Longer Problems
29
B. p* = 200(15) - 5(15)(35) + 100(35) = $3,875
C. The interpretation of l = ∂p/∂k is that by increasing the
production requirement that x + y = 5 by one unit, then
maximum profits will increase by $25. Alternatively, by reducing
production requirements by one unit, then maximum profits will
decline by $25.
D. The first-order conditions for profit maximization in the
unconstrained case are
∂p/∂x = 200 - 5y = 0
∂p/∂y = - 5x + 100 = 0
y* = 40
x* = 25
x* + y* = 65
b* = 200(25) - 5(25)(40) + 100(40) = $4,000
WSG2 7/7/03 4:32 PM Page 30