Module 1: Marginal analysis and single variable calculus 6. Answers to ...

[Pages:7]1 May 2019 Module 1: Marginal analysis and single variable calculus

6. Answers to selected exercises

Exercise 1: Profit-maximizing firm

(a) Invert the demand function

p 26q .

(b) Then revenue is R(q) qp(q) q(26q) so profit is

(q)

R(q)

C(q)

20q

5 2

q2

F

(c) The derivative is

(q) 205q 5(4q) .

Thus profit has a critical point at q 4 .

The derivative (the slope of the graph of the profit function) is strictly positive for q q and strictly negative for q q . Thus (q) is maximized at q 4

(d)

(q)

20(4)

5 2

(16)

F

40 F

.

Thus the firm makes a positive profit if F 40 . In case (ii) it is better off not producing so maximized profit is zero.

Exercise 2: Profit maximization

(a) (x) 2x3 9x2 12x2

Therefore marginal profit is

M (x) 6x2 18x12 6(x2 3x2) 6(x1)(x2) . Thus the critical values for this function are x 1 and x 2 .

The marginal profit function is depicted below.

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1 May 2019 Module 1: Marginal analysis and single variable calculus

Fig 1: Marginal profit function

Note that the slope of the graph of the profit function, (x) , is negative on (0,1) and positive on (1,2) . Therefore the profit is decreasing on (0,1) and increasing on (1,2) Thus x 1 is a local minimum. Also (x) , is negative for all x 2 thus the profit function is decreasing for all x 2 . Thus x 2 is a local maximum.

We have shown that the profit function is decreasing on (0,1) , increasing on (1,2) and decreasing again for all higher x . Therefore the graph of the profit function is as depicted

below.

Fig 2: Profit function

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1 May 2019 Module 1: Marginal analysis and single variable calculus

As is readily checked, profit is negative at x 2 . Thus it is not profitable for the firm to enter

the market.

Exercise 3: Input Choice

(a) Revenue is R p2q

Output increases with input according to q 10z1/2 . Therefore we can substitute and write revenue as a function of z .

R 10p2z1/2.

This is called the revenue product of the input.

Total cost is C p1z . Therefore profit is

(z) p2q p1z 10p2z1/2 p1z

(c)

(z)

5p2z1/2

p1

,

(z)

5 2

p2z3/2

0

The slope is strictly positive for sufficiently small z therefore the profit maximizing input is

strictly positive. At the maximum, z , (z) 5p2z 1/2 p1 . Therefore

z1/2 5p2 and so z (5p2 )2 .

p1

p1

Since the slope is strictly decreasing, z is the unique maximizer.

Substituting into production function, q(z) 50 p2 . Substituting into the profit function, p1

(z) 25p22 . p1

Exercise 4: Price taking firm

(a) (q) R(q) C(q) 6q 11q2 2q3 1 q4

2

4

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1 May 2019 Module 1: Marginal analysis and single variable calculus

(b)-(d) (q) 611q6q2 q3 (1q)(2q)(3q).

The critical points are {1,2,3}.

(e) The derivative is positive on [0,1] and [2,3]. It is negative on [1,2] and [3,] .

Thus {1,3} are local maxima and 2 is a local minimum.

(f)

If you compute the profit at each local maximum you will find that in each case

(q)

2

1 4

.

So the function has two global maxima.

(g) Excel graph of profit function

Exercise 5: Derivative of a cubic Applying the rules of multiplication,

f (ab) (ab)3 a3 3a2b3ab2 b3

Therefore

f (ab) f (a) 3a2b3ab2 b3 (3a2 3abb2)b .

Hence

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1 May 2019 Module 1: Marginal analysis and single variable calculus

f (a b) f (a) 3a2 3abb2 b

The limit as b 0 is therefore 3a2 . This holds for all a . Therefore f (x) 3x2 .

Exercise 6:

Derivative of

f

(x)

x2

1 x2

f

(ab)

f

(a)

1 (a b)2

1 a2

a2 (a b)2 (a b)2a2

a2

a2 2abb2 (a b)2a2

2ab b2 (a b)2a2

(2a (a

b)b b)2 a2

Therefore

f

(a

b) b

f

(a)

2a b (a b)2a2

The derivative is the limit as

b 0 . Therefore

f

(a)

2 a3

2a3

.

Exercise 7: Discriminating monopoly

(a)

The demand function is

q1

30

1 2

p1 .

To invert this function we rearrange the terms as

follows:

1 2

p1

30 q1

.

Therefore p1(q1) 602q1 .

(b) Revenue is R(q1) q1p1(q1) 60q1 2q12 and total cost is 20q1. So profit is

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1 May 2019 Module 1: Marginal analysis and single variable calculus

(q1) R(q1) C(q1) 40q1 2q12 .

Marginal profit is therefore

M (q1) 404q1.

Note that marginal profit is decreasing. Therefore the profit function is concave. The critical

point q1 satisfies

(q1 ) 404q1 0 .

Therefore q1 10 and so p1(q1 ) 602q1 40 . As is easily checked, maximized profit is 200.

(c) Rearranging the demand function yields the European demand price function

p2(q2) 33q2 .

Total export revenue is therefore R2(q2) q2 p2(q2) 33q2 q22 . Total profit is

(q1,q2) R1(q1) R2(q2) 20(q1 q2)

[R1(q1) 20q1][R2(q2) 20q2] .

Thus to maximize total profit we need to maximize each bracketed expression. The first is USA profit and the second is European profit. We have already solved for the USA maximizing price. European profit is

2 R2(q2) 20q2 13q2 q2

d2 dq2

13 2q2

.

The critical point satisfies

d2 dq2

13 2q2

0

so

q2

6.5 . Therefore

p2(q2 ) 33q2

26.5

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1 May 2019 Module 1: Marginal analysis and single variable calculus

Exercise 8: Common price

(a)

With a common price

p

the two demand functions are

q1

30

1 2

p

and

q2 33 p then

total

demand

is

q q1

q2

63

3 2

p.

Inverting this equation, the demand price function is

p

42

2 3

q

.

(6-1)

Total profit is then

(q)

p(q)q 20q

22q

2 3

q2

.

(6-2)

Marginal profit is therefore

(q)

22

4 3

q

.

This is strictly decreasing so the profit function is strictly concave. The critical point is q* 33 . 2

Substituting into (6-1) p* 31 .

Remark: It is actually easier to solve for the common price that maximizes total profit

expressed as a function of p .

(c) Substituting into (6-2), maximized profit is

(q*) 363 181.5 . 2

From the previous question, if the firm only sells in the domestic market it has a profit of 200. Thus exporting is no longer profitable.

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