However, in this particular example we know for certain that zero slope corresponds to the maximum value of the function.. 9.2 Second-order condition for a maximum In the example in Sect
Trang 19 Unconstrained optimization
Learning objectives
After completing this chapter students should be able to:
• Find the maximum or minimum point of a single variable function by differenti-ation and checking first-order and second-order conditions
• Use calculus to help find a firm’s profit-maximizing output
• Find the optimum order size for a firm wishing to minimize the cost of holding inventories and purchasing costs
• Deduce the comparative static effects of different forms of taxes on the output of
a profit-maximizing firm
9.1 First-order conditions for a maximum
Consider the total revenue function
TR= 60q − 0.2q2
This will take an inverted U-shape similar to that shown inFigure 9.1 If we ask the question
‘when is TR at its maximum?’ the answer is obviously at M, which is the highest point on the curve At this maximum position the TR schedule is flat To the left of M, TR is rising and has a positive slope, and to the right of M, the TR schedule is falling and has a negative slope At M itself the slope is zero
We can therefore say that for a function of this shape the maximum point will be where its slope is zero This zero slope requirement is a necessary first-order condition for a maximum.
Zero slope will not guarantee that a function is at a maximum, as explained in the next section where the necessary additional second-order conditions are explained However, in this particular example we know for certain that zero slope corresponds to the maximum value of the function
InChapter 8, we learned that the slope of a function can be obtained by differentiation
So, for the function
TR= 60q − 0.2q2
slope=dTR
dq = 60 − 0.4q
© 1993, 2003 Mike Rosser
Trang 20
£
M
q
Figure 9.1
The slope is zero when
60− 0.4q = 0
60= 0.4q
150= q
Therefore TR is maximized when quantity is 150
Test Yourself, Exercise 9.1
1 What output will maximize total revenue if TR= 250q − 2q2?
2 If a firm faces the demand schedule p = 90 − 0.3q how much does it have to sell
to maximize sales revenue?
3 A firm faces the total revenue schedule TR= 600q − 0.5q2
(a) What is the marginal revenue when q is 100?
(b) When is the total revenue at its maximum?
(c) What price should the firm charge to achieve this maximum TR?
4 For the non-linear demand schedule p = 750 − 0.1q2what output will maximize the sales revenue?
9.2 Second-order condition for a maximum
In the example in Section 9.1, it was obvious that the TR function was a maximum when its slope was zero because we knew the function had an inverted U-shape However, consider the function inFigure 9.2(a) This has a slope of zero at N, but this is its minimum point not its maximum In the case of the function in Figure 9.2(b) the slope is zero at I, but this is neither a maximum nor a minimum point
The examples in Figure 9.2 clearly illustrate that although a zero slope is necessary for a function to be at its maximum it is not a sufficient condition A zero slope just means that
the function is at what is known as a ‘stationary point’, i.e its slope is neither increasing nor decreasing Some stationary points will be turning points, i.e the slope changes from positive
© 1993, 2003 Mike Rosser
Trang 30
y
(a)
(b) y
x
x
T S
N
I
Figure 9.2
to negative (or vice versa) at these points, and will be maximum (or minimum) points of the function
In order to find out whether a function is at a maximum or a minimum or a point of inflexion
(as in Figure 9.2(b) ) when its slope is zero we have to consider what are known as the second-order conditions (The first-second-order condition for any of the three forms of stationary point is
that the slope of the function is zero.)
The second-order conditions tell us what is happening to the rate of change of the slope of the function If the rate of change of the slope is negative it means that the slope decreases as the variable on the horizontal axis is increased If the slope is decreasing and one is at a point where the actual slope is zero this means that the slope of the function is positive slightly
to the left and negative slightly to the right of this point This is the case inFigure 9.1 The slope is positive at Y, zero at M and negative at Z Thus, if the rate of change of the slope of
a function is negative at the point where the actual slope is zero then that point is a maximum This is the second-order condition for a maximum Until now, we have just assumed that
a function is maximized when its slope is zero if a sketch graph suggests that it takes an inverted U-shape From now on we shall make this more rigorous check of the second-order conditions to confirm whether a function is maximized at any stationary point
It is a straightforward exercise to find the rate of change of the slope of a function We
know that the slope of a function y = f(x) can be found by differentiation Therefore if we differentiate the function for the slope of the original function, i.e dy/dx, we get the rate of
change of the slope This is known as the second-order derivative and is written d2y/ dx2
© 1993, 2003 Mike Rosser
Trang 4Example 9.1
Show that the function y = 60x − 0.2x2satisfies the second-order condition for a maximum
when x= 150
Solution
The slope of this function will be zero at a stationary point Therefore
dy
x = 150
Therefore the first-order condition for a maximum is met when x is 150.
To get the rate of change of the slope we differentiate (1) with respect to x again, giving
d2y
dx2 = −0.4
This second-order derivative will always be negative, whatever the value of x Therefore, the second-order condition for a maximum is met and so y must be a maximum when x is 150.
In the example above, the second-order derivative did not depend on the value of x at the
function’s stationary point, but for other functions the value of the second-order derivative may depend on the value of the independent variable
Example 9.2
Show that TR is a maximum when q is 18 for the non-linear demand schedule.
p = 194.4 − 0.2q2
Solution
TR= pq = (194.4 − 0.2q2)q = 194.4q − 0.2q3
For a stationary point on this cubic function the slope must be zero and so
dTR
dq = 194.4 − 0.6q2= 0
194.4 = 0.6q2
324= q2
18= q When q is 18 then the second-order derivative is
d2TR
dq2 = −1.2q = −1.2(18) = −21.6 < 0
© 1993, 2003 Mike Rosser
Trang 5Therefore, second-order condition for a maximum is satisfied and TR is a maximum when
q is 18 (Note that in this example the second-order derivative−1.2q < 0 for any positive value of q.)
Test Yourself, Exercise 9.2
Find stationary points for the following functions and say whether or not they are at their maximum at these points
1 TR= 720q − 0.3q2
2 TR= 225q − 0.12q3
3 TR= 96q − q 1.5
4 AC= 51.2q−1+ 0.4q2
9.3 Second-order condition for a minimum
By the same reasoning as that set out in Section 9.2 above, if the rate of change of the slope
of a function is positive at the point when the slope is zero then the function is at a minimum This is illustrated inFigure 9.2(a) The slope of the function is negative at S, zero at N and positive at T As the slope changes from negative to positive, the rate of change of this slope must be positive at the stationary point N
Example 9.3
Find the minimum point of the average cost function AC= 25q−1+ 0.1q2
Solution
The slope of the AC function will be zero when
dAC
0.2q = 25q−2
q3= 125
q = 5 The rate of change of the slope at this point is found by differentiating (1), giving the second-order derivative
d2AC
dq2 = 50q−3+ 0.2
= 50
125+ 0.2 when q = 5
= 0.4 + 0.2 = 0.6 > 0
© 1993, 2003 Mike Rosser
Trang 6Therefore the second-order condition for a minimum value of AC is satisfied when q is 5 The actual value of AC at its minimum point is found by substituting this value for q into
the original AC function Thus
AC= 25q−1+ 0.1q2= 25
5 + 0.1 × 25 = 5 + 2.5 = 7.5
Test Yourself, Exercise 9.3
Find whether any stationary points exist for the following functions for positive
values of q, and say whether or not the stationary points are at the minimum values of
the function
1 AC= 345.6q−1+ 0.8q2
2 AC= 600q−1+ 0.5q 1.5
3 MC= 30 + 0.4q2
4 TC= 15 + 27q − 9q2+ q3
5 MC= 8.25q
9.4 Summary of second-order conditions
If y = f(x) and there is a stationary point where dy
dx = 0, then
(i) this point is a maximum if d
2y
dx2 <0
(ii) this point is a minimum ifd
2y
dx2 >0
Strictly speaking, (i) and (ii) are conditions for local maximums and minimums It is possible,
for example, that a function may take a shape such as that shown inFigure 9.3 This has no
true global maximum or minimum, as values of y continue towards plus and minus infinity
as shown by the arrows Points M and N, which satisfy the above second-order conditions for maximum and minimum, respectively, are therefore just local maximum and minimum points However, for most of the examples that you are likely to encounter in economics any local maximum (or minimum) points will also be global maximum (or minimum) points and
so you need not worry about this distinction If you are uncertain then you can always plot a function using Excel to see the pattern of turning points
If d2y/ dx2= 0 there may be an inflexion point that is neither a maximum nor a minimum, such as I inFigure 9.2(b) To check if this is so one really needs to investigate further, looking at the third, fourth and possibly higher order derivatives for more complex polynomial functions However, we will not go into these conditions here In all the economic applications given
in this text, it will be obvious whether or not a function is at a maximum or minimum at any stationary points
Some functions do not have maximum or minimum points Linear functions are obvi-ous examples as they cannot satisfy the first-order conditions for a turning point, i.e that
dy/dx = 0, except when they are horizontal lines Also, the slope of a straight line is always
© 1993, 2003 Mike Rosser
Trang 7N M
–y
y
Figure 9.3
a constant and so the second-order derivative, which represents the rate of change of the slope, will always be zero
Example 9.4
InChapter 5we considered an example of a break-even chart where a firm was assumed to have the total cost function TC= 18q and the total revenue function TR = 240 +14q Show
that the profit-maximizing output cannot be determined for this firm
Solution
The profit function will be
π = TR − TC
= 240 + 14q − 18q
= 240 − 4q
Its rate of change with respect to q will be
dπ
There is obviously no output level at which the first-order condition that dπ /dq = 0 can
be met and so no stationary point exists Therefore the profit-maximizing output cannot be determined
End-point solutions
There are some possible exceptions to these first- and second-order conditions for maximum and minimum values of functions If the domain of a function is restricted, then a maximum or
© 1993, 2003 Mike Rosser
Trang 8minimum point may be determined by this restriction, giving what is known as an ‘end-point’
or ‘corner’ solution In such cases, the usual rules for optimization set out in this chapter will not apply For example, suppose a firm faces the total cost function (in £)
TC= 45 + 18q − 5q2+ q3
For a stationary point its slope will be
dTC
However, if we try using the quadratic equation formula to find a value of q for which (1)
holds we see that
q =−b ±
√
b2− 4ac
√
102− 4 × 18 × 3
10±√−116 6
We cannot find the square root of a negative number and so no solution exists There is no
turning point as no value of q corresponds to a zero slope for this function.
However, if the domain of q is restricted to non-negative values then TC will be at its minimum value of £45 when q= 0 Mathematically the conditions for minimization are not met at this point but, from a practical viewpoint, the minimum cost that this firm can ever face is the £45 it must pay even if nothing is produced This is an example of an end-point solution
Therefore, when tackling problems concerned with the minimization or maximization of economic variables, you need to ask whether or not there are restrictions on the domain of the variable in question which may give an end-point solution
Test Yourself, Exercise 9.4
1 A firm faces the demand schedule p = 200 − 2q and the total cost function
TC= 2
3q3− 14q2+ 222q + 50
Derive expressions for the following functions and find out whether they have
maximum or minimum points If they do, say what value of q this occurs at and
calculate the actual value of the function at this output
(a) Marginal cost
(b) Average variable cost
(c) Average fixed cost
(d) Total revenue
(e) Marginal revenue
(f) Profit
2 Construct your own example of a function that has a turning point Check the second-order conditions to confirm whether this turning point is a maximum or a minimum
3 A firm attempting to expand output in the short-run faces the total product of labour schedule TPL = 24L2− L3 At what levels of L will (a) TPL, (b) MPL, and (c) APLbe at their maximum levels?
© 1993, 2003 Mike Rosser
Trang 94 Using your knowledge of economics to apply appropriate restrictions on their domain, say whether or not the following functions have maximum or minimum points
(a) TC= 12 + 62q − 10q2+ 1.2q3
(b) TC= 6 + 2.5q
(c) p = 285 − 0.4q
9.5Profit maximization
We have already encountered some problems involving the maximization of a profit function
As profit maximization is one of the most common optimization problems that you will encounter in economics, in this section we shall carefully work through the second-order condition for profit maximization and see how it relates to the different intersection points of
a firm’s MC and MR schedules
Consider the firm whose marginal cost and marginal revenue schedules are shown by MC and MR in Figure 9.4 At what output will profit be maximized?
The first rule for profit maximization is that profits are at a maximum when MC= MR However, there are two points, X and Y, where MC= MR Only X satisfies the second rule for profit maximization, which is that MC cuts MR from below at the point of intersection This corresponds to the second-order condition for a maximum required by the differential calculus, as illustrated in the following example
Example 9.5
Find the profit-maximizing output for a firm with the total cost function
TC= 4 + 97q − 8.5q2+ 1/3q3
and the total revenue function
TR= 58q − 0.5q2.
X
0
MC
MR
Y
D
q
£
Figure 9.4
© 1993, 2003 Mike Rosser
Trang 10First let us derive the MC and MR functions and see where they intersect
dq = 97 − 17q + q2
(1)
Therefore, when MC= MR
97− 17q + q2= 58 − q
(3− q)(13 − q) = 0
These are the two outputs at which the MC and MR schedules intersect, but which one satisfies the second rule for profit maximization? To answer this question, the problem can
be reformulated by deriving a function for profit and then trying to find its maximum Thus, profit will be
π = TR − TC = 58q − 0.5q2− (4 + 97q − 8.5q2+ 1/3q3)
= 58q − 0.5q2− 4 − 97q + 8.5q2− 1/3q3
= −39q + 8q2− 4 − 1/3q3 Differentiating and setting equal to zero
dπ
0= 39 − 16q + q2
(5)
Equation (5) is the same as (3) above and therefore has the same two solutions, i.e q = 3 or
q = 13 However, using this method we can also explore the second-order conditions From (4) we can derive the second-order derivative
d2π
dq2 = 16 − 2q
When q= 3 then d2π/ dq2= 16 − 6 = 10 and so π is a minimum.
When q= 13 then d2π/ dq2= 16 − 26 = −10 and so π is a maximum.
Thus only one of the intersection points of MR and MC satisfies the second-order conditions for a maximum and corresponds to the profit-maximizing output This will be where MC cuts MR from below We can prove that this must be so as follows:
By differentiating (1) we get
slope of MC= dMC
dq = −17 + 2q
© 1993, 2003 Mike Rosser