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CHAPTER 2
OPTIMAL DECISIONS USING
MARGINAL ANALYSIS
The following table shows the five topic sections of this chapter and the associated study
guide problems that pertain to each topic section.
Section Topic
1
A Simple Model of the Firm
Problems M1-M5.
2
Marginal Analysis
Problems M6-M7, S1, L1.
3
Marginal Revenue and Marginal Cost
Problems M8-M13, S2-S4, L2-L5.
4
Sensitivity Analysis
Problems M14-M19, L6-L9.
5
Appendix: Calculus and Optimization Techniques
Problems M20, S5-S8, L10-L11.
Multiple Choice
M1
According to the model of the firm, management’s main objective is to maximize
a.
Revenue.
b.
Return on Investment.
c.
Profit.
d.
Market share in targeted segments.
e.
Earnings growth.
M2
According to the simple model of the firm, management can predict
a.
Both revenues and costs with certainty.
b.
Costs with certainty, but revenues imprecisely.
c.
Its output level with certainty, but the market-clearing price imprecisely.
d.
Neither revenues nor costs very precisely.
e.
Revenues and costs with certainty only in the short run (next 3 months).
M3
A firm’s total cost function is given by: C = 1,000 + 15Q. At an output of 200 units,
a.
Total cost is $4,000 and marginal cost is $3,000.
b.
Total cost is $3,000 and fixed cost is $1,000.
c.
Total cost is $4,000 and marginal cost is 15.
d.
Average total cost is constant.
e.
Answers c and d are both correct.
10 Managerial Economics Study Guide
M4
A firm’s demand curve is described by the equation: Q = 200 – 4P. The inverse demand
curve is:
a.
P = 200 – 4Q.
b.
P = 50 - .25Q.
c.
P = 50 - .5Q.
d.
P = 200 - .25Q.
e.
None of the above answers is correct.
M5
A firm’s price and cost equations are given by P = 200 - 2Q and C = 1,000 + 40Q,
respectively. Therefore,
a.
The firm’s revenue is: R = 200 – 2Q2.
b.
The firm’s profit is:  = -1000 + 160Q – 2Q2
c.
The firm’s revenue is: R = 200Q – 2Q2.
d.
At Q = 50 units, the firm’s total profit is 2,500.
e.
Answers b and c are both correct.
M6
At its current output level, a firm’s marginal profit is negative. Therefore, it should
a.
Increase output until M = 0
b.
Reduce output because MR > MC.
c.
Reduce output because MR < MC.
d.
Shut down production to avoid continuing losses.
e.
Reduce both its output and its price.
M7
A firm’s profit equation is given by:  = -100 + 160Q – 20Q2. Therefore,
a.
The firm’s fixed cost is 100.
b.
M = 160 – 20Q.
c.
The firm’s profit-maximizing output is Q = 4.
d.
M = 160 – 40Q.
e.
Answers a, c, and d are all correct.
M8
For a downward sloping demand curve, the associated marginal revenue curve
a.
Coincides with the demand curve.
b.
Lies below and parallel to the demand curve.
c.
Has the same price intercept but a steeper slope than the demand curve.
d.
Is positive for all levels of sales.
e.
None of the above answers is correct.
M9
A firm can sell as much output as it wishes at the fixed price, P = $10 per unit. Then,
a.
The revenue curve has the usual inverted U-shape.
b.
To maximize profit, the firm will produce at full capacity.
c.
MR = 10 - 2Q.
d.
Marginal revenue is constant and equal to $10.
e.
Answers a and c are both correct.
From https://testbankgo.eu/p/Solution-Manual-for-Managerial-Economics-8th-Edition-by-Samuelson
M10
In order to maximize profit, management should set output such that
a.
Marginal profit is zero.
b.
Marginal profit is maximized.
c.
Marginal revenue is zero.
d.
Marginal revenue equals marginal cost.
e.
Answers a and d are both correct.
M11
A firm’s price and cost equations are given by P = 200 - .2Q and C = 1,000 + 40Q,
respectively. Therefore, its profit maximizing level of output is
a.
Q = 500 units.
b.
Q = 400 units.
c.
Q = 300 units
d.
Q = 600 units.
e.
None of the above answers is correct.
M12
A firm’s price equation is given by: P = 200 - 20Q. Then, the firm’s
a.
Revenue-maximizing output is 6 units.
b.
Revenue-maximizing price is $100.
c.
Profit-maximizing price is $100.
d
The firm’s profit-maximizing output is 3 units.
e.
Answers a and d are both correct.
M13
A firm’s cost equation is given by C = 200 + 10Q + 2Q2. At Q = 20, marginal cost is
a.
$1,200
b.
1,200/20 = $6
c.
$1,000
d.
1,000/20 = $5
e.
$90
M14
If fixed costs increase, what will happen to marginal revenue and marginal cost?
a.
Both will increase.
b.
Both will decrease.
c.
Marginal cost will increase; marginal revenue will be unchanged.
d.
Both will be unchanged.
e.
Marginal cost will increase; marginal revenue will decrease.
M15
Suppose that the demand for a product increases. What is the most likely effect on the
firm’s price and quantity?
a.
Both will increase.
b.
Both will be unchanged.
c.
Both will decrease.
d.
Management can choose to hold one of them constant and increase the other.
e.
Price will decrease, quantity will increase.
12 Managerial Economics Study Guide
M16
Suppose that the firm’s marginal cost decreases. What is the effect on the firm’s optimal
price and quantity?
a.
Both will increase.
b.
Both will be unchanged.
c.
Price will decrease, quantity will increase.
d.
There is not enough information to determine how price and quantity will change.
e.
Price will increase, quantity will decrease.
M17
Suppose that demand decreases. What is the most likely effect on the marginal revenue
and marginal cost curves?
a.
Marginal revenue will decrease, marginal cost will not change.
b.
Marginal revenue and marginal cost will both decrease.
c.
Neither will change.
d.
Marginal revenue will not change, marginal cost will decrease.
e.
Marginal revenue will increase, marginal cost will decrease.
M18
According to the text, what was the nature of the conflict between Burger King and the
local franchisees?
a.
Burger King wanted the franchisees to maximize profit, but the franchisees
wanted to maximize revenue.
b.
Burger King wanted to maximize revenue for itself, but wanted the franchisees to
maximize profit.
c.
Burger King wanted the franchisees to maximize revenue, but the franchisees
wanted to maximize profit.
d.
Burger King wanted the franchisees to minimize cost of sales, but the franchisees
wanted to maximize revenue.
e.
Burger King wanted the franchisees to maximize output, but the franchisees
wanted to maximize revenue.
M19
The fact that Amazon profits by selling a predictable number of e-books for each Kindle
it sells implies that
a.
It earns additional MR for each Kindle sold.
b.
It should raise the price of Kindle so as to avoid cannibalizing e-book sales.
c.
It can raise the price of the Kindle and still increase the quantity of Kindles it
sells.
d.
It should discount the price of the Kindle.
e.
Answers a and d are both correct.
M20
If afirm’s profit is given by:  = -Q3+18Q2 - 60Q -100, then its optimal output is
a.
Q=2
b.
Q = 10
c.
Q = 14
d.
A maximum does not exist. Profit is unbounded.
e.
None of these answers is correct.
From https://testbankgo.eu/p/Solution-Manual-for-Managerial-Economics-8th-Edition-by-Samuelson
Short Problems and Questions
S1
Suppose that the firm’s marginal cost is $500 per unit and that demand is as given in the
following situations. In each case, find the profit-maximizing level of output.
a.
P = 1,500 - 25Q.
b.
P = 2,000 - 50Q.
c.
Q = 3,000 - 5P.
S2
Suppose that a firm faces the demand curve, P = 30 - 1.5Q, where P denotes price in
dollars and Q denotes total unit sales. The cost equation is C = 28 + 12Q.
a.
Determine the firm’s profit-maximizing output and price.
b.
Suppose that there is a change in the labor contract so that the cost equation
becomes C = 15 + 16Q + .25Q2. Determine (and explain) the resulting effect on
the firm’s output and price.
c.
Suppose that the firm sells in a competitive market and faces the fixed price, P =
$21. For the cost function in part b, find the firm’s new optimal quantity.
S3
Suppose that a farmer produces grain that sells in a highly competitive market where the
price is $4 per unit. The cost equation is given by C = 360 + .01Q2.
a.
Determine the optimal quantity for the farmer to produce.
b.
Now suppose that the market price falls to $3.60 per unit. What is the new
optimum? Verify the result that a drop in price leads to a reduction in output.
S4
Compute average total cost (C/Q) and marginal cost for the following cost functions.
a.
C = 80Q.
b.
C = 20 + 40Q.
c.
C = 100 + 30Q + 5Q2.
S5
What is the second derivative and why is it important for optimization problems?
S6
In each case below, find the profit-maximizing level of output. Verify that each is a
maximum by checking the second derivative.
a.
 = -50 + 100Q - 10Q2.
b.
 = -20 + 150Q - 2Q3.
c.
 = -40 - 180Q + 39Q2 - 2Q3.
S7
Compute the derivatives of the following functions:
a.
y = 4x1.5.
b.
y = 1/x.
S8
Compute the appropriate partial derivatives for the following functions.
a.
Q = 12K.5L.5.
b.
Q = PaYb.
14 Managerial Economics Study Guide
Longer Problems and Discussion Questions
L1
Define marginal analysis and explain how management uses it in decision making.
L2
A small liberal arts college is trying to predict enrollment for the next academic year. The
vice president for business states that enrollment has tended to follow a pattern described
by E = 18,000 – .5P, where E denotes total enrollment and P is yearly tuition.
a.
If the school sets tuition at $20,000, how many students can it expect to enroll?
b.
If the school seeks to enroll 6,000 students, what tuition should it charge?
c.
If the school wants to maximize total tuition revenue, what tuition should it
charge?
d.
As the vice president for business, what tuition would you recommend? Why?
L3
Land Grant College has a successful football team, and sells tickets to students, alumni,
and the public. Experience has shown that attendance has followed the demand
relationship: Q = 80,000 - 5,000P where Q is attendance per game and P is the ticket
price in dollars. Current capacity of the football stadium is 50,000 people.
a.
If Land Grant charges $7 per ticket, predict attendance.
b.
The athletic director wants to maximize revenue in order to pay for other athletic
programs at Land Grant. What is the appropriate ticket charge?
c.
Because the stadium is to be substantially repaired, capacity next year will be
limited to 35,000 seats. Given this limitation, what ticket price should Land Grant
set?
L4
A firm faces the inverse demand curve, P = 20 - .5Q and cost function, C = 25 + 2Q + Q2.
Prepare a table listing quantities from 0 to 12 and corresponding prices, revenues, costs,
and profits. In the final two columns, list marginal revenues and marginal costs. Verify
that the MR = MC rule results in maximum profit.
L5
State the basic rule for maximizing profit in production. Carefully explain why this isan
appropriate rule to follow.
L6
John Deere sells tractors and other agricultural equipment in competition U.S. and
international manufacturers. Over a one-year period, the U.S. dollar as depreciated about
25% vis-à-vis other foreign currencies. How would this affect Deere’s demand curve? In
turn, what would be the ultimate effect on Deere’s output and average sales price?
L7
The Dutch Peddler is a small bicycle repair shop, specializing in minor repairs and tuneups. The direct, out-of-pocket cost of a spring tune-up is $18, and the price is currently set
at $30. Dutch Peddler has noticed that its number of tune-ups has increased in each of the
past four years, while its tune-up price has remained unchanged. The owner of Dutch
Peddler is an old college friend of yours, and asks you for some free advice on how to
improve earnings at the shop. In particular, she is considering a price increase. What
would you advise?
L8
Keith Johnson, manager of the Campus BookStore, has to make a decision. Recently, a
sales rep from a clothing manufacturer offered to produce sweatshirts with the school’s
From https://testbankgo.eu/p/Solution-Manual-for-Managerial-Economics-8th-Edition-by-Samuelson
logo. The quoted cost is $10 per sweat (including shipping) plus $1,600 to prepare the
patterns for producing the logo. Keith has heard from other bookstore managers at a
recent convention that school-labeled products sell very well in college bookstores, so he
is seriously considering adding this line of clothing to boost earnings.
a.
Demand for sweats is: P = 40 - .02Q. Determine the profit-maximizing price and
quantity for Campus BookStore.
b.
Due to limited shelf space, Johnson must limit the number of sweats that he
stocks to 80% of the profit-maximizing quantity found in part a. What is the
“marginal cost” associated with this constraint? Explain.
c.
Now suppose demand falls to: P = 20 - .02Q. (Costs remain the same as before.)
Is it still possible to earn a profit on sweats? Explain.
L9
Suppose that a firm employs a sales force that is compensated on a commission system.
The firm pays a 10% commission on the first $300,000 of gross sales booked by a seller
and a 15% commission on sales over $300,000.
a.
Is there a conflict in objectives between the sales force and management of the
firm? Explain.
b.
The use of an increasing commission rate provides a strong incentive for agents to
increase gross sales. When would a company not want to encourage increased
sales by its agents? Explain.
L10
New Car Motors sells cars and trucks in a rapidly growing suburb. Based on past trends,
the sales manager estimates that the inverse demand equation for cars is: PC = 40 -.05QC
where QC is the number of cars sold per year and PC is the price per car (in thousandsof
dollars). The price equation for trucks is: PT = 30 - .1QT. In turn, NCM pays $10 thousand
for each car from the manufacturer and $8 thousand for each truck.
a.
Determine the profit-maximizing outputs of cars and trucks for NCM.
b.
Answer the questions in part a, supposing that capacity is limited to a total of 260
vehicles per year.
L11
Suppose that Q (output) depends upon the use of inputs K (capital) and L (labor)
according to the production function:
Q = K.5L.5
The firm has allocated $1,000 per week to buy capital and labor, where K costs $4 per
unit and L costs $5 per unit. The firm seeks to maximize output subject to its $1,000
budget. How much of each input should the firm buy? What is its maximum level of
output? Use the method of Lagrangian multipliers to address these questions.
SOLUTIONS
Multiple Choice
M1
M2
M3
M4
c
a
c
b
This is a simplifying assumption.
Total cost is C = 1,000 + (15)(200) = $4,000 and MC = dC/dQ = $15 per unit.
16 Managerial Economics Study Guide
M5
M6
M7
M8
M9
M10
M11
M12
e
c
e
c
d
e
b
b
M13
M14
M15
M16
M17
M18
M19
M20
e
d
a
c
a
c
e
b
In perfect competition, the price equation is simply P = $10, implying MR = $10.
The conditions MR = MC and M = 0 are equivalent.
Setting MR = MC implies 200 - .4Q = 40, or Q = 400.
To maximize revenue, set MR = 0. Thus, 200 - 40Q = 0, implying Q = 5 and
P = $100.
MC = dC/dQ = 10 + 4Q. Thus, at Q = 20, MC = $90.
A change in fixed cost has no effect on marginal cost or marginal revenue.
M = -3Q2 + 36Q – 60 = -3(Q -2)(Q – 10). The maximum is at Q = 10.
Short Problems and Questions
S1
a.
b.
c.
S2
a.
b.
c.
S3
a.
b.
To maximize profit, set MR = MC. We know that revenue is: R = PQ so
R = (1,500 - 25Q)(Q) = 1,500Q - 25Q2, implying: MR = 1,500 - 50Q.
Setting MR = MC = 500 implies 1,500 - 50Q = 500 or 1,000 = 50Q. Thus, the
optimal quantity is Q* = 20.
P = 2,000 - 50Q implies MR = 2,000 - 100Q. Again, setting MR = MC, we have:
2,000 - 100Q = 500 or 1,500 = 100Q. Thus,Q* = 15.
Rearranging Q = 3,000 - 5P, we have: 5P = 3,000 – Q, or P = 600 -.2Q.
Setting MR = MC, we have 600 - .4Q = 500. Thus, Q* = 250.
= R - C = (30 - 1.5Q)(Q) – [28 + 12Q]. After simplifying, this becomes:
 = 18Q - 1.5Q2 - 28. Therefore, M = d/dQ = 18 - 3Q. Setting M = 0implies:
18 - 3Q = 0, or Q* = 6. In turn, P* = 30 - (1.5)(6) = $21.
Equivalently, if we set MR = MC, we have MR = 30 – 3Q = 12, so Q* = 6.
From the new cost equation, we have: MC = dC/dQ = 16 + .5Q. Again, setting
MR = MC implies: 30 - 3Q = 16 + .5Q. Thus, 14 = 3.5Q or Q* = 4. Because MC
has increased (relative to part a), the firm’s optimal quantity declines. In turn,
P* = 30 - (1.5)(4) = $24.
With the price fixed at $21, it follows that R = 21Q and MR = dR/dQ = 21.
Setting MR = MC implies 21 = 16 + .5Q, or Q* = 10. With MR constant (rather
than downward sloping), it pays for the firm to increase output.
The farmer’s profit is:  = R - C = 4Q - .01Q2 – 360. Thus, M = d/dQ =
4 - .02Q. Setting M = 0 implies: 4 - .02Q = 0, or Q* =200.
At P = $3.60, profit is:  = 3.6Q - .01Q2 – 360. Thus, M = d/dQ = 3.6 - .02Q.
Setting M = 0 implies: 3.6 - .02Q = 0, or Q* = 180. This verifies that a drop in
price leads to a reduction in output.
From https://testbankgo.eu/p/Solution-Manual-for-Managerial-Economics-8th-Edition-by-Samuelson
S4
a.
b.
c.
S5
The second derivative is found by twice differentiating a given function, that is, by taking
the first derivative then taking the first derivative of that. For example, marginal profit
(the rate of change of profit) is found by taking the derivative of the profit function. By
taking the derivative of marginal profit, we find the second derivative of the profit
function. For instance, if = -50 + 20Q - 2Q2, then we have: M = d/dQ = 20 - 4Q and
d2/dQ2 = d(M)/dQ = -4.
The sign of the second derivative determines whether a function is at a maximum (if
the sign is negative) or a minimum (if the sign is positive). In this example, setting M =
0 implies Q*= 5. Since the second derivative is negative (-4) at this output, we know that
this is a maximum.
S6.
a.
b.
c.
With C = 80Q, AC = 80Q/Q = 80 and MC = dC/dQ = 80.
With C = 20 + 40Q, AC = 20/Q + 40 and MC = 40.
With C = 100 + 30Q + 5Q2, AC = 100/Q + 30 + 5Q and MC = 30 + 10Q.
M = d/dQ = 100 - 20Q = 0 implies Q*= 5.
In turn, d2/dQ2 = -20, so we have found a maximum.
M = d/dQ = 150 - 6Q2 = 0 implies Q* = 5. In turn, d/dQ2 = -12Q, which is
negative at Q* = 5, so we have found a maximum.
M = d/dQ = -180 + 78Q - 6Q2 = 0. Dividing each side by 6 and rearranging, we
get : Q2 – 13Q + 30 = 0, which factors into (Q - 3)(Q – 10) = 0. Thus, the solution
is Q* = 3 or Q* = 10. In turn, d2 /dQ2 = 78 - 12Q. This is positive at Q* = 3 and
negative at Q* = 10. The first output is a minimum and the latter is a maximum.
S7
a.
b.
dy/dx = 6x.5 = 6 x .
y = 1/x = x-1. Therefore, dy/dx = -x-2 = -1/x2.
S8
a.
b.
For Q = 12K.5L.5, Q/K = 6K-.5L.5 and Q/L =6K.5L-.5.
For Q = PaYb, Q/P = aPa-1Yb and Q/Y= bPaYb-1.
Longer Problems and Discussion Questions
L1
Marginal analysis is the process of considering a small change in a decision and
determining whether the given change will improve the ultimate objective. Management
uses marginal analysis to identify an optimal decision – that is, a decision that maximizes
a given objective (usually profit).For instance, a firm might try charging a slightly
different price and monitoring the effect on unit sales, revenues, costs, and profits.
Alternatively, it might change its mix of inputs (capital, labor, and so on) in seeking to
minimize the cost of producing a given level of output.
L2
a.
b.
c.
From the demand equation, E = 18,000 - (.5)(20,000) = 8,000 students.
Rearranging the demand equation (.5P = 18,000 – E) give the price equation
P = 36,000 – 2E. For E = 6,000, we find: P = $24,000.
Tuition revenue is maximized when MR = 0. Revenue = PE = (36,000 - 2E)E.
18 Managerial Economics Study Guide
d.
L3
a.
b.
c.
L4
Thus, R = 36,000E - 2E2,implyingMR = dR/dE = 36,000 - 4E = 0. Consequently,
E =9,000 students and tuition, P =36,000 – (2)(9,000) = $18,000 per year.
The recommended tuition depends on several factors. One is the cost of operating
the school, especially the marginal cost of an additional student. Another is the
objective of the school. Colleges are interested in attracting high-quality students
and faculty and in enhancing their academic reputations. Typically, liberal arts
colleges do not try to maximize profit, but they do pay attention to enrollment and
revenue. By offering financial aid, they also offer different tuition charges to
different students.
Q = 80,000 - (5,000)($7) = 80,000 - 35,000 = 45,000 people.
Start by rearranging the demand equation: Q = 80,000 - 5,000P becomes
5,000P = 80,000 – Q, or P = 16 – Q/5,000. To maximize revenue, the athletic
director sets MR = 0. From the price equation, it follows that MR = 16 – 2Q/5,000
or MR = 16 – Q/2,500. Setting MR = 0 implies 16 – Q/2,500 = 0, or Q = 40,000.
In turn, the appropriate ticket price is P = 16 – 40,000/5,000 = $8. Maximum
revenue is R = ($8)(40,000) = $320,000.
To fill the reduced stadium capacity (35,000 seats), the appropriate price is:
P = 16 - 35,000/5,000 = 16 - 7 = $9. This generates less revenue ($315,000) than
in part b
Revenues are computed using: R = 20Q - .5Q2. In turn, MR = dR/dQ = 20 – Q. From the
cost equation, C = 25 + 2Q + Q2, we have: MC = dC/dQ = 2 + 2Q.
After completing the table, we find the profit-maximizing output to be Q* = 6. At this
output, profit is 29 and MR = MC = 14. The same result can be found algebraically by
setting MR = MC implying: 20 - Q = 2 + 2Q. Therefore, Q = 6.
Q
1
2
3
4
5
6
7
8
9
10
11
12
L5
P
19.5
19
18.5
18
17.5
17
16.5
16
15.5
15
14.5
14
R
19.5
38
55.5
72
87.5
102
115.5
128
139.5
150
159.5
168
C
28
33
40
49
60
73
88
105
124
145
168
193

-8.5
5
15.5
23
27.5
29
27.5
23
15.5
5
-8.5
-25
MR
19
18
17
16
15
14
13
12
11
10
9
8
MC
4
6
8
10
12
14
16
18
20
22
24
26
The firm’s profit maximizing level of output occurs when the additional revenue from
selling an extra unit just equals the extra cost of producing it, that is, when MR = MC.
This rule is appropriate because it focuses attention on the effects of a change in current
activities, in particular, the effects of expanding or contracting production. If marginal
revenue exceeds marginal cost, this means the additional revenue from producing and
selling a unit exceeds the cost of doing so. This adds net profit to the firm (that is, it
From https://testbankgo.eu/p/Solution-Manual-for-Managerial-Economics-8th-Edition-by-Samuelson
increases the firm’s total profit or it reduces its loss). If marginal revenue is less than
marginal cost, then increasing the activity implies a net reduction in profit.
In a multi-products firm, if the products are assumed to be independent of one another,the
firm should follow this rule for each of the products. In this case, maximum total profit is
achieved by maximizing profit product by product.
L6
This dollar depreciation makes imports of foreign goods to the U.S. more expensive in
dollar terms, while making Deere’s exports less expensive to the rest of the world in
terms of their foreign currencies. Both effects increase demand for Deere’s goods causing
its demand curve to shift outward. In responsive to the positive demand shift, Deere’s
new optimal decision (at the intersection of a shifted MR and an unchanged MC) calls for
increased Q* and P*. Deere should partially increase its prices and also increase its
planned sales.
L7
This problem is a bit subtle, because demand is not known with precision. However, we
know some things that can help us make a recommendation about a price change.
(1) The goal of the manager is to increase profits, which gives an adviser a clearly stated
goal for the decision-maker.
(2) Though price has been constant, quantity has steadily increased over the years. Thus,
demand is increasing (the demand curve is shifting to the right). If the $30 price was once
appropriate, a higher price in response to increased demand is probably appropriate now.
(An increase in MC over time would also warrant an increase in price.)
(3) Overall, it seems that a small price increase (say, to $35 or $40) will likely increase
total revenue and profit (even though it will slow the growth in volume). If the manager
pursues this course, she should carefully track unit sales to measure the willingness of
customers to pay the increased price.
L8
a.
b.
c.
L9
a.
Johnson finds the optimal quantity and price of sweats by setting MR = MC.
From the price equation, P = 40 - .02Q, it follows that MR = 40 - .04Q. The cost
of an additional sweatshirt is $10. Thus, 40 - .04Q = 10 or Q* = 750 sweats.
The optimal price is: P = 40 - (.02)(750) = $25.
Suppose Johnson is limited to ordering (.8)(750) = 600 sweats. At this quantity,
marginal revenue is: MR = 40 - (.04)(600) = $16, while MC = $10. Thus, the store
would earn a marginal profit of $6 by ordering and selling an extra sweatshirt.
This lost profit (for unsold shirts 601 up to 750) is the “cost” of the limitation to
600 sweats.
If demand falls to P = 20 - .02Q then, MR becomes 20 - .04Q. Setting MR = MC,
we find 20 - .04Q = 10, or Q = 250 shirts. In turn, P* = $15. The store’s profit on
the sweats falls to:  = (15 - 10)(250) - 1,600 = -$350. With the fall in demand,
ordering the sweats becomes unprofitable.
No, there is not necessarily a conflict. Clearly, because of the commission
structure, the sales force has a strong incentive to increase sales Top management
also wants to increase sales, as long as the additional revenue of the sales exceeds
the additional cost of production and sales commission.
20 Managerial Economics Study Guide
L10
b.
Beyond some level, a company might want to discourage additional sales in two
cases: 1) When the increased (net) revenue is insufficient to cover the increased
cost of commissions. This is possible and suggests that management should
examine the commission scheme and perhaps lower commission rates or increase
prices. 2) When the firm faces a constraint in production, making an increase in
production very costly or impossible. In this case, any increase in sales is either
infeasible (leaving disappointed customers) or not worth the cost.
a.
The profit from cars is: C = R - C = 40QC - .05Q2C - 10QC.
Setting M = 0 implies: 30 - .1QC = 0, or QC = 300 cars.
The profit from trucks is: T = R - C = 30QT - .1Q2T - 8QT.
Setting M = 0implies: 22 - .2QT = 0, or QT = 110 trucks.
NCM is limited to 260 vehicles (instead of the desired 410 vehicles in part a).
The relevant constraint is: QC + QT = 260. Before, NCM attained maximum profit
by setting MC = 0 and MT = 0. Now, the best it can do is to set MC = MT.
Equating marginal profits implies: 30 - .1QC = 22 - .2QT. Solving this equation
and QC + QT = 260 simultaneously, we find: QC = 200 cars and QT = 60 trucks.
b.
L11
The objective is to maximize Q, subject to the constraint of spending no more than
$1,000 for inputs. The Lagrangian is: £ = K.5L.5 + z(1,000 - 4K - 5L).
Compute the partial derivatives, and set them equal to zero:
£/K = .5K-.5L.5 - 4z = 0
£/dL = .5K.5L-.5 - 5z = 0
£/z = 100 - 4K - 5L = 0
1) We multiply the first partial by 5 and the second partial by 4 so as to cancel the z term:
2.5K-.5L.5 - 20z = 0 and 2K.5L-.5 - 20z = 0. Therefore, 2.5K-.5L.5 = 2K.5L-5.
2) Next, we multiply both sides by K.5L.5. Therefore, 2.5L = 2K so that K = 1.25L.
3) Finally, substitute this into the constraint: 4K + 5L = 1,000, implying
4(1.25L) + 5L = 1,000 or 10L = 1,000. Thus, L* = 100 and K* = (1.25)(100) = 125.
Total output is (125).5(100).5 = 111.8 units.