9781118041581

For example, suppose we have y x/(8 x). Then

Let’s derive an expression for marginal profit (denoted by M) applying

these calculus rules to our profit function:

From Rule 4, we know we can proceed term by term. From Rule 2, the deriva-

tive of the first term is 2. According to Rule 3, the derivative of the second term

is .2Q. From Rule 1, the derivative of the third term is zero. Thus,

Notice the elegance of this approach. By substituting specific values of Q, we

can find marginal profit at any desired level of output. For instance, at Q 5,

we find that M 2 (.2)(5) 1; at Q 12, M.4; and so on.

To determine the firm’s optimal output level, we set M0. Thus,

Solving this equation for Q, we find Q 10. This confirms that the profit-

maximizing level of output is 10 thousand units.

THE SECOND DERIVATIVE In general, one must be careful to check that a

maximum, not a minimum, has been found. In the previous example, the

graph makes it clear that we have found a maximum. But suppose the profit

expression is more complicated: say,

[2A.2]

Figure 2A.2 shows the associated profit graph. Notice that there are two quan-

tities at which the slope is zero: one is a maximum and the other is a minimum.

It would be disastrous if we confused the two. Taking the derivative of the profit

function, we find

Substitution confirms that marginal profit is zero at the quantities Q 2 and

Q 10. The graph shows that Q  2 minimizesprofit, whereas Q  10 maximizes

profit.

There is a direct way to distinguish between a maximum and a minimum.

At a maximum, the slope of the profit function changes from positive to zero

Md/dx3.6Q.3Q^2 6.

1.8Q^2 .1Q^3 6Q10.

2 .2Q0.

Md/dx 2 .2Q.

2Q.1Q^2 3.6.

dy/dx[1#(8x) 1 #(x)]/(8x)^2 8/(8x)^2.

66 Appendix to Chapter 2 Calculus and Optimization Techniques

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