MC, AC - fdslive.oup.com€¦ · Chapter 8: Economic applications of functions and derivatives Answers to further student exercises ... MC, by definition, is the derivative of TC,

Renshaw: Maths for Economics 4e 1 Chapter 8: Economic applications of functions and derivatives

Answers to further student exercises

© Geoff Renshaw, 2016. All rights reserved.

Exercise WS8.1

1. (a) 0 25 20TC . q

(b) 0.25dTC

MCdq

. 20

0.25TC

ACq q

, or 10.25 20AC q

0

5 10 15 20 25 30 35 40 45 50q

MC, AC

0.25

MC

AC

Ex W8.1 question 1(a)

0

10

20

30

40

50

5 10 15 20 25 30 35 40 45 50q

TC

0.25 20TC q

Ex WS8.1 question 1(a)




1.(c) Fixed costs, by definition, are the constant term in the TC function. Because

MC, by definition, is the derivative of TC, MC is independent of fixed costs.

This may be seen in (b) above, where the fixed costs of 20 do not appear in

the MC expression. However since average cost is simply total cost divided

by output, AC is not independent of fixed costs. This may also be seen in (b)

above, where fixed costs (20) appear in the AC expression, divided by output.

Thus AC always exceeds MC as long as there are any fixed costs.

1.(d) From (b) we have AC – MC = 20 20

(0.25 ) 0.25 .q q

Thus AC exceeds MC by

20

q, which we can interpret as fixed costs per unit of output. Because

20

q

approaches zero as q gets larger and larger, AC approaches closer and

closer to MC as q increases (because fixed costs are spread over a larger

and larger output). The sketch graphs in (b) above show this.

2. (a) 23 5 300TC q q

(b) 6 5MC q . 300

3 5AC qq

, or 13 5 300AC q q

Ex W8.1 question 2(a)

0

100

200

300

400

500

600

700

800

900

1000

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15q

TC

Ex WS8.1 question 2(a)




(c) Show that AC is at its minimum when q = 10, and that MC = AC at this output.

Minimum AC where 0dAC

dq and

2

20

d AC

dq . Here, we have

23 300 0.dAC

qdq

So 2

3003

q 2 300

1003

q 100 10q (ignoring the negative root).

Also

2

3

2 3

600600

d ACq

dq q

, which is positive when q is positive, as we assume to be

the case.

When q = 10, 6 5MC q = 65, and 300

3 5AC qq

= 65, so MC = AC when

AC at its minimum. Alternatively, if we set MC = AC we get 6 5q =

3003 5q

q . This simplifies to 2 100q , with a positive root of q = 10. (As

usual we ignore the negative solution.)

(d) Sketch the graphs of the MC and AC functions, on the same axes.

3003 5AC q

q

Ex W8.1 question 2(d)

0

50

100

150

200

250

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15q

MC, AC

6 5MC q

Ex WS8.1 question 2(d)




3. Given the short run total cost function 3 22.2 16 48 150TC q q q

(a) Show that average cost is at its minimum when q = 5. We have

2 1502.2 16 48

TCAC q q

q q , or 2 12.2 16 48 150

TCAC q q q

q

Min AC occurs where 0dAC

dq and

2

20

d AC

dq . Here,

2

2

1504.4 16 150 4.4 16

dACq q q

dq q

. When q = 5,

2

1504.4(5) 16 22 16 6 0

5

dAC

dq . Also,

23

2 3

3004.4 300 4.4

d ACq

dq q

, which is

positive when q = 5. So when q = 5, 0dAC

dq and

2

20

d AC

dq as is required for a

minimum of AC. Also, when q = 5, 2 1502.2(5) 16(5) 48 53

5AC

(b) Find the output at which marginal cost is at its minimum.

26.6 32 48dTC

MC q qdq

. So 13.2 32.dMC

qdq

Setting this equal to zero gives

13.2 32 0q , with solution 2.42q (to 2 dp). We have 2

213.2

d MC

dq which is positive

when 2.42q so this point is a minimum.

(c) Show that MC = AC when AC is at its minimum.

From (a) above, when AC is at its minimum, q = 5 and AC = 53. But when q = 5, 26.6(5) 32(5) 48 53MC

(d) Sketch the graphs of the MC and AC functions, on the same axes.

We know from above that the MC curve has an intercept of 48 on the vertical axis,

has its minimum at 2.42q , and cuts the AC curve at q = 5, AC = MC = 53. This

information is sufficient to permit us to sketch the graph of the MC curve, below.




Similarly, we know from above that the AC is asymptotic to the vertical axis, because

its last term, 150

q, goes to infinity as q approaches zero. We also know that the AC

curve has its minimum at 5q , AC = 53, and cuts the MC curve at this point. This

information is sufficient to permit us to sketch the graph of the AC curve, below.

Note: AC is asymptotic to the vertical axis.

MC

AC


0

20

40

60

80

100

120

140

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15q

MC, ACEx WS8.1 question 3(d)




(e) We know that MC minimum is at q = 2.42. Because MC is measured by

the tangent to the TC curve, we know that at this point the tangent is at its flattest

(least positive slope). That is, the tangents drawn on both sides of this point are

steeper. This helps us to sketch the curve in the vicinity of q = 2.42. We also know

that MC = AC at q = 5. This means that the dotted line from the origin (a "ray" from

the origin), which measures AC, is also tangent to the TC curve and thus measures

MC too. This helps us sketch the curve in the neighbourhood of q = 5. To the right

of q = 5, the curve rises continuously. The y intercept is 150.

(f) The point of minimum MC is the point at which the slope of the TC curve is at

its minimum, meaning that at that point a small increase in output can be

achieved at lowest additional cost. This may be because the machinery that

produces this output is operating at its highest efficiency at that point. There

is no great economic significance to this point. Note, incidentally, that in the

range of output between q = 2.42 and q = 5, MC is rising but AC is falling.

Ex W8.1 question 3(e)

0

200

400

600

800

1000

1200

1400

1 2 3 4 5 6 7 8 9 10

q

TC

Ex WS8.1 question 3(e)




Exercise WS8.2

1.(a) Given the inverse demand function 0.1 50p q ,

The TR and MR functions, as functions of q, are:

2( 0.1 50) 0.1 50TR pq q q q q (note, a quadratic function). So

0.2 50dTR

MR qdq

(b) Find the price and quantity at which total revenue is maximised.

TR is maximised when 0dTR

MRdq

and 2

20.

d TR

dq So we need

0.2 50 0dTR

MR qdq

, which gives q = 250. We also have 2

20.2

d TR

dq which is

negative, so we have a maximum of TR at

q = 250. Using the inverse demand function , we find p = 25 when q =250.




(c) Sketch the graphs of the inverse demand function and the marginal revenue

functions on the same axes, indicating your solution on the graph. Sketch the

graph of the total revenue function.

Ex W8.2 question 1(c)

-80

-60

-40

-20

0

20

40

60

0 50 100 150 200 250 300 350 400 450 500 550 600q

p, MR

-4000

-2000

0

2000

4000

6000

8000

50 100 150 200 250 300 350 400 450 500 550 600

q

TR

20.1 50TR q q

0.1 50p q

0.2 50MR q

Ex WS8.2 question 1(c)




2. Consider the demand function 23 324q p

(a) Find the total and marginal revenue functions.

First we have to find the inverse demand function (which expresses p as a function

of q). This is a little difficult as the demand function is quadratic. Given 23 324q p , by elementary operations we get 23 324p q 2 1

3108p q

0.513

( 108)p q (the inverse demand function). Using this, 0.513

( 108) .TR qp q q

(Note, we cannot multiply out the brackets because of the power 0.5).

Now we can find MR as the derivative of 0.513

( 108) .TR q q . Using both the product

rule and the function of a function rule, we find this derivative as

0.5 0.51 1 13 3 3

(0.5)( 108) ( ) ( 108) (1)dTR

MR q q qdq

which simplifies a little to

MR = 0.5 0.51 1 16 3 3

( 108) ( 108)q q q

(b) Find the price and quantity at which TR is at its maximum. (Hint: It's easier to

start by finding the price rather than the quantity at which TR is maximised).

For the moment we will ignore the hint and try to find max TR by setting our

expression above for MR equal to zero and solving the resulting equation. This

gives

MR = 0.5 0.51 1 16 3 3

( 108) ( 108) 0q q q

This looks very difficult to solve but the trick is to multiply both sides by 0.513

( 108)q

(This multiplication of course leaves the right hand side of the equation unchanged.)

This multiplication gives us:

0.5 0.5 0.5 0.51 1 1 1 16 3 3 3 3

( 108) ( 108) ( 108) ( 108) 0MR q q q q q

By adding the powers (and recalling that anything raised to the power zero equals 1)

this simplifies to

1 16 3

( 108) 0MR q q




which is easily solved to give q = 216. We then find p by substituting q = 216 into the

inverse demand function, giving 0.513

( 108)p q = 0.513

( (216) 108)p = 6

Alternatively, following the hint, we start from the demand function 23 324q p .

From this, we get 2 3( 3 324) 3 324TR pq p p p p . (Note that this gives TR as a

function of price, which is perfectly valid, but differs from the conventional definition

which gives TR as a function of quantity, as used earlier in this question.)

We can then differentiate this to get 29 324dTR

MR pdp

. (Again, this is a function

of p which is not the usual definition of MR, but it is perfectly valid.) We then set this

equal to zero to find the p at which TR is at its maximum. Thus 29 324 0p . The

positive solution to this quadratic equation is p = 6 (we ignore the negative solution).

We then find q by substituting p = 6 into the demand function, giving 23 324q p

23(6) 324 216q .

(c) Sketch the graph of the demand function, the MR and the TR functions (the

last is a little difficult!)

To sketch the graphs we proceed as follows.

(i) The demand function. The demand function is 23 324q p . We see that this is

a quadratic function and its graph is therefore a parabola, sketched below. (The q

intercept is obviously 324; the p intercept is found by setting q = 0 and solving for p.)

q

10.39 p

324

0

Demand function 23 324q p




The inverse demand function is then simply this same curve, but with axes

interchanged (and negative values of p discarded):

(ii) The MR function. From (a) above we have:

MR = 0.5 0.51 1 16 3 3

( 108) ( 108)q q q

To find its intercept on the vertical axis we set q = 0, giving 0.5108 10.39 (to 2 d.p.).MR (Note that this is the same as the intercept of the inverse

demand function, above). Since we know that MR = 0 when q = 216, this gives us

the intercept on the q axis. However, this is not quite enough to enable us to sketch

the MR function; we also need to know whether it is linear, concave or convex. We

can get this information by looking at the sign of the second derivative, which will be

zero, negative or positive according to whether the curve is linear, concave or

convex. Finding the first and second derivatives of 0.5 0.51 1 1

6 3 3( 108) ( 108)MR q q q is a lengthy and tedious task, but requires

nothing more than a careful application of the rules of differentiation. The required

differentiation is done in the annex below, where we show that the first derivative and

second derivatives are both negative. This tells us that the MR curve is downward

sloping and concave from below. On reflection this concavity is not surprising, given

that the inverse demand curve is also concave (see sketch above). So when the

inverse demand function and the MR function are sketched on the same axes, they

look like this:

p

10.39

q 324 0

Inverse demand function 0.51

3( 108)p q




Annex to question 2(c): First and second derivatives of (A) inverse demand function

and (B) marginal revenue function.

(A) We have inverse demand function 0.513

( 108)p q . Using the function of a

function rule, we get 0.5 0.51 1 1 1 13 3 6 3 6

1(0.5)( 108) ( ) ( )( 108) ( )

dpq q

dq p

since

0.5 0.513

( 108)p q u ,

This is negative when, as we assume, p is positive. So the inverse demand function

is negatively sloped.

To find the second derivative, we start from 0.51 16 3

( )( 108)dp

qdq

, from which, using

the function of a function rule, we get 2

1.5 1.51 1 1 1 1 16 3 3 36 3 362 3

1( )( 0.5)( 108) ( ) ( 108)

d pq q

dq p

. This is negative when,

as we assume, p is positive. So the inverse demand function is concave. The

shape of the inverse demand function in the sketch above is therefore correct.

(B) We have 0.5 0.51 1 16 3 3

( 108) ( 108)MR q q q . As in (A) we can simplify this by

defining 1 13 3

108 from which .du

u qdq

Then we have 0.5 0.516

MR qu u .

Therefore: 1.5 0.5 0.51 16 6

( )( 0.5 ) ( ) 0.5dMR du du

q u u udu dq dq

(Note that the expression in square brackets is the derivative of 0.51 16 3

( 108)q q ,

216

p,MR

10.39

q 324 0

Inverse demand function 0.51

3( 108)p q

Marginal revenue function 0.5 0.51 1 1

6 3 3( 108) ( 108)MR q q q




found by applying both the product rule and the function of a function rule.) This

simplifies to

1 12 21.5 0.5 31 1 1 1

36 3 36 3( ) ( ) ( )

dMRqu u q u u

dq

Since we have 0.5 0.513

( 108)p q u , the above simplifies to

1 136 33

1dMR q

dq pp

As this is negative since p is assumed positive, this shows that the MR function is

downward sloping.

To find the second derivative, we start from

1.5 0.51 136 3

( )dMR

qu udq

from which

2

2.5 1.5 1.51 1 136 36 32

1.5 ( ) 0.5d MR du du

q u u udq dqdq

(Note again that the expression in square brackets is the derivative of 1.5136

qu ,

found by applying both the product rule and the function of a function rule.) This

simplifies to

2

2.5 1.51 172 122

d MRqu u

dq

Again, since we have 0.5 0.513

( 108)p q u , the above simplifies to

2

1 172 122 5 3

1 1( )

d MRq

dq p p

This is negative since we assume p to be positive, so the MR function is concave

from below. So the MR function is concave, and the shape of the MR function in the

sketch above is therefore correct.

(iii) The TR function: 0.513

( 108) .TR q q This is relatively easy to sketch. Given that

TR pq we know that TR = 0 when either p or q is zero. From the demand function

we see that q = 324 when p = 0. So the intercepts of the TR function on the q axis




are at q = 0 and q = 324. We also know that TR reaches its maximum at q = 216. At

this point, 0.5 0.51 13 3

( 108) 216( (216) 108) 1296.TR q q With these three pieces of

information we can sketch the function:

3. The inverse demand function for a good is 543

2q

p

(a) Sketch the inverse demand function, indicating clearly the intercepts on the p

and q axes (if they exist) and/or the asymptotes (if they exist). (Hint: you

should recognise this function as a rectangular hyperbola.)

Given 543

2q

p

we see that when q = 0, p = 18 – 2 = 16. And when p = 0, 543

2q

,

from which q = 24. Also, we recognise the function as a rectangular hyperbola.

There is a discontinuity at q = –3. As q increases without limit, p approaches a

limiting value of –2. This information is sufficient to draw the sketch below with

reasonable accuracy.

(b) Find the total and marginal revenue functions, and the values of p and q that

maximise total revenue.

Given 543

2q

p

we have 54

32

q

qTR q

, from which

2 2

( 3)54 54 (1) 162

( 3) ( 3)2 2

q q

q qMR

.

For max TR we need MR = 0 2

162

( 3)2 0

q 2( 3) 81q 2 6 72 0q q .

The two roots to this quadratic equation are q = 6 and q = –12. It is left to you as an

exercise to find 2

2d TR

dqand show that this is negative at q = 6, which is therefore the

216

TR

1296

q 324 0

Total revenue function 0.51

3( 108)TR q q




maximum of TR. (We ignore the q = -12 root as quantity cannot be negative.) When

q = 6, from the inverse demand function p = 4. Therefore max TR is pq = 4(6) = 24.

(c) Sketch the total and marginal revenue functions, indicating clearly the

relationship between the two and their relationship with the inverse demand

function.

(i) The TR function. First we will find the intercepts. Given , we have TR = 0 when 54

32 0

q

qq

54 2 ( 3)q q q 22 48 0q q . The roots of this quadratic are q = 0

and q = 24, so these are the intercepts on the q axis. We also know that the TR

function reaches its max at q = 6, p = 4. From this we can sketch the TR function

below.

(ii) The MR function. From (b) above we know that 0MR when q = 6, so this is the

intercept on the q axis. We find the intercept on the MR axis by setting q = 0, giving 162

92 16.MR We can show that the MR curve is downward sloping and convex

from below, as follows.




We have 2

2162

( 3)2 162( 3) 2

qMR q

Using the function of a function rule, we find

the derivative as 3

3

324324( 3)

( 3)

dMRq

dq q

, which is negative when q is positive,

showing that the MR curve is negatively sloped. (We could also have found this

derivative using the quotient rule.) Differentiating again gives us 2

4

2 4

972972( 3)

( 3)

d MRq

dq q

, which is positive when q is positive, showing that the

MR curve is convex from below. See sketches below.

Note: As shown algebraically above, both inverse demand function and MR function

intercept vertical axis at p = MR = 16 (not shown in sketch above).

Ex W8.2 question 3

-1

0

1

2

3

4

5

6

7

2 4 6 8 10 12 14 16 18 20 22 24q

p, MR

p

MR

0

5

10

15

20

25

30

2 4 6 8 10 12 14 16 18 20 22 24

q

TR

54

32

q

qTR q

54

32

q

qp q

2162

( 3)2

qMR

Ex WS8.2 question 3




4.(a) We assume that the inverse demand function is ( )p f q where ( )f q denotes

some unspecified function of q. The slope of the inverse demand function is

'( )dp

f qdq

, where '( )f q means whatever we get when we differentiate the

unspecified function ( )f q .

By definition we have TR pq , so by substituting for p we can write ( )TR f q q .

Marginal revenue is the derivative of this. Using the product rule, we get

'( )(1) ( )dTR

MR f q qf qdq

which from above is the same thing as

dp

MR p qdq

Looking at this expression we see that MR depends on p, q and the slope of

the inverse demand function .dp

dq Since q is not a constant, one way in which

MR can be constant is if 0.dp

dq This means that the inverse demand function

is horizontal (see sketch below), and therefore that p is constant. From our

expression for MR above, we then have

a constant.MR p

(b) Is there a way in which MR can be constant but not equal to price? For this to

occur, we need dp

p qdq

to be constant, this time with dp

dq not equal to zero.

We can get this result by assuming the inverse demand function to have the

form kp Aq where A and k are constants (parameters). Then we have

1kdpAkq

dq

(power rule), and therefore:

1( ) (1 )k kdpMR p q p q kAq p kAq p kp p k

dq

From this, we see that MR can be constant and not equal to p, provided k = 1.

p, MR

p = MR

q 0

Inverse demand function

( ) with 0dp

p f qdq




Then MR = 0 (a constant). This means that the inverse demand function is

1 Ap Aq

q

, which is a rectangular hyperbola (see sketch below). Note that

in this case total revenue is given by 1( )TR pq Aq q A . Thus total revenue

is constant (whatever the values of p and q) and given by the parameter A.

Exercise WS8.3

1. A firm's short run total cost function is 20.5 500TC q q . The firm sells in a

perfectly competitive market and the ruling price is p = 61.

(a) Find the most profitable level of output, and the profits at that output.

The TR function is 62 .TR pq q Therefore the profit function is

2 261 (0.5 500) 0.5 60 500TR TC q q q q q . Therefore

60.d

qdq

For max profit we set this equal to zero and solve, with solution q =

60. Also 2

21.

d

dq

This is negative when q = 60. So we have a maximum of at q

= 60 because 0d

dq

and

2

20

d

dq

.

When q = 60, 2105(60) 60(60) 500 1300

p

q 0

Inverse demand function

A

pq

This implies MR = 0




(b) Does the firm produce at minimum average cost? Explain your answer.

The AC function is 500

0.5 1TC

AC qq q

or 10.5 1 500AC q q . Therefore

2

2

5000.5 500 0.5

dACq

dq q

We set this equal to zero and solve,

2 500

0.5q q = 31.62 to 2 decimal places. (We ignore the negative root.)

Also 2

2 3

10000

d

dq q

when q >0. So we have a minimum of AC at q = 31.62 because

0dAC

dq and

2

20

d AC

dq .

Ouput at which AC is minimised (q = 31.62) is lower in this case than output at which

is maximised (q = 60). But this is not surprising as the former requires 0dAC

dq

while the latter requires 0d

dq

; there is no reason why the value of q that satisfies

one of these equations should also satisfy the other. Another way of explaining this

point is to note that profits are maximised when marginal cost equals marginal

revenue; average cost is irrelevant.




(c)

Ex W8.3 question 1

0

1000

2000

3000

4000

5000

6000

7000

10 20 30 40 50 60 70 80 90 100 110

q

TC, TR

-500

0

500

1000

1500

10 20 30 40 50 60 70 80 90 100 110

q

0

20

40

60

80

100

120

140

160

10 20 30 40 50 60 70 80 90 100 110q

MC, MR

Profit

TC

TR

MR

MC

MR2

-500

q

TR2

Ex WS8.3 question 1




(c) Sketch the graphs of total cost and total revenue with the same axes, and

do the same with marginal cost and marginal revenue. Sketch the graph of the profit

function. Indicate in all diagrams the equilibrium values of the variables.

In sketching these graphs (see above) we use the following information. The total

cost function is 20.5 500TC q q which is quadratic with an intercept of 500 on the

TC axis. Its first derivative is 1dTC

qdq

which is positive when q is positive. Its

second derivative is 2

21

d TC

d q which is positive. So the TC curve is upward sloping

and convex from below, as drawn above.

Further, the derivative 1dTC

qdq

gives us the marginal cost function, so the graph of

MC is linear with an intercept of 1 and a slope of 1

Since price is constant at p = 61, total revenue is TR = pq = 61q, a linear function

pasing through the origin with a slope of 91. Marginal revenue is 61dTR

MRdq

; its

graph is a horizontal line (independent of q). (Note that MR and p are equal).

The profit function is 260 0.5 500q q which you should recognise as a quadratic

which is an inverted U-shape because the coefficient of x2 is negative. Its intercept

is -500 (= profits when output is zero). From (a) above we know its maximum is at q

= 60.

2.(a) If we re-solve question 1 above with p = 45, the new profit maximising output

level is q = 44 and profits are 468.

(b) In the graphs for question 1, the fall in price reduces the slope of the TR

function, giving the new curve TR2. The resulting new profit function is 2,

with its maximum at q = 44 and profits of 468.Because MR = p when price is

constant, the fall in price from 61 to 45 causes the MR curve to drop from 61

to 45 (see graphs above). The new MR curve cuts the MC curve at q = 44,

the new profit maximizing output.




3.(a) Repeating question 2 above we find that when the ruling market price falls to

33 the most profitable output is q = 32 with profits 12.

(b) If the firm produces zero output, its TR is zero and its costs are 500 (= its fixed

costs, which have to be met whatever output is produced, including zero

output). Therefore its profits are –500. Therefore it is better for the firm to

produce a positive output, even at a loss (negative profits) provided losses are

less than 500. In other words the firm will only cease production if the price

falls so low that losses are greater than 500 at the profit maximizing output.

To find this price, we begin by re-writing the profit function with an unspecified

price, p. From question 1 above, we get

2 2(0.5 500) 0.5 ( 1) 500TR TC pq q q q p q

where p is a constant (parameter). Therefore ( 1).d

q pdq

For max profit

we set this equal to zero and solve, giving 1q p . Thus when profits are

maximized, we will have 1q p . If we assume that profits are indeed

maximized (that is, given any price the firm always produces the profit

maximizing output), we can substitute p–1 in place of q in the profit function,

and get

20.5( 1) ( 1)( 1) 500p p p which simplifies to

20.5( 1) 500p

Looking at this expression, we can see that there are three caes:

(i) if p > 1, profits are greater than –500, and it is worthwhile for the firm to

continue to produce. (For example, if p = 2, profits are 20.5(2 1) 500 499.5. )

(ii) If p = 1, profits are –500, and it appears that the firm will be indifferent

whether it ceases production or not. However this overlooks the condition

derived above, that 1q p . This means that when p = 1, the profit

maximizing output is q = 0; that is, the firm will cease production.

(iii) A puzzle arises when p < 1. For example, if p = 0.5, 20.5( 0.5) 500 0.125 500 499.875 . As –499.875 is greater than

–500 (a loss of 499.875 is smaller than a loss of 500), it appears that the firm

will continue to produce rather than shutting down. This is puzzling, because




it appears that profits when p = 0.5 (-499.875) are higher than when p = 1

(–500). We would not expect a lower price to yield a higher profit. The

resolution of this paradox appears when we again recall from above that profit

maximization requires 1q p . This means that, when p < 1, q < 0. But

output cannot be negative, so we must assume that when p < 1, q = 0. That

is, the firm ceases production and its losses are 500.

So we conclude that when p = 1, the profit maximizing output is q = 0; that is,

the firm will cease production. When p < 1, the profit maximizing output is

negative, but as this is impossible the firm will again cease production. So

only if p > 1 will the firm produce any output.

(Not an easy question!)

4.(a) If we repeat question 1(a) with the new TC function we find the most profitable

output is q = 17. Maximized profits are –66.5. The relevant graphs are below,

where we see that the TC curve now lies entirely above the TR curve, so the

firm makes losses at any level of output. However q = 17 is the point where

the TC curve is closest to the TR curve, hence losses are minimized (= profits

maximized).




(b)

Ex W8.3 question 4

0

1000

2000

3000

10 20 30 40 50

q

TC, TR

-500

0

500

1000

1500

10 20 30 40 50

q

0

20

40

60

80

100

120

140

160

10 20 30 40 50

q

MC, MR

MR

TC

MC

TR

Ex WS8.3 question 4




4. A firm's short run total cost function is 20.75 8 300TC q q . The firm is a

monopolist and the inverse demand function for its product is 201 2 .p q

(a) Find the most profitable level of output, and the profits at that output.

Solving this exactly as in previous questions in this exercise, we get q = 35.09 (to 2

dp), = 3069 (to 4 significant figures) and p = 130 approx.

(b) When maximizing profit, does the firm produce at minimum average cost?

The average cost function is 10.75 8 300AC q q . So

2

2

3000.75 300 0.75

dACq

dq q

. Setting this equal to zero and solving for q gives q =

20 as output at which AC is minimized. (The 2nd order conditions for a minimum are

also satisfied, because 2

3

2 3

600600

d ACq

dq q

, which is positive as required for a min.)

So the firm produces to the right of the minimum point on its AC curve.

(c) Find the break-even points.

We find these by setting TR = TC and solving for q. This gives 22.75 193 300 0.q q

This quadratic has solutions q = 1.59 and q = 68.59.

(d) Sketch the graphs of the inverse demand function, marginal revenue and

marginal cost function with the same axes, showing the equilibrium price, MR

and MC, and output. See below.

(e) Sketch the graphs of total cost and total revenue with the same axes, showing

the equilibrium output and profits. Sketch the graph of the profit function,

showing the equilibrium output and profits. Show the break-even points on

both graphs. See below.




Graph for question 5(d)

MC

p

MR


-20

30

80

130

180

10 20 30 40 50 60 70

q

p, MR. MC




Graphs for question 5(e)

Ex W8.3 question 6(e)

0

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2000

3000

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6000

10 20 30 40 50 60 70

q

TC, TR

TC

TR

Max profit

-500

0

500

1000

1500

2000

2500

3000

3500

10 20 30 40 50 60 70

q

Break-even points

Max profit

Break-even points

Documents

MC, AC - fdslive.oup.com€¦ · Chapter 8: Economic applications of functions and derivatives Answers to further student exercises ... MC, by definition, is the derivative of TC,