Derivatives as Rates of Change: Learn It 4

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Derivatives as Rates of Change: Learn It 4

Rate of Change Applications Cont.

Changes in Cost and Revenue

In addition to analyzing motion along a line and population growth, derivatives are useful in analyzing changes in cost, revenue, and profit. The concept of a marginal function is common in the fields of business and economics and implies the use of derivatives.

The marginal cost is the derivative of the cost function.
The marginal revenue is the derivative of the revenue function.
The marginal profit is the derivative of the profit function, which is based on the cost function and the revenue function.

Marginal Cost, Marginal Revenue and Marginal Profit

If $C(x)$ is the cost of producing $x$ items, then the marginal cost $MC(x)$ is
$MC(x)=C^{\prime}(x)$ .
If $R(x)$ is the revenue obtained from selling $x$ items, then the marginal revenue $MR(x)$ is
$MR(x)=R^{\prime}(x)$ .
If $P(x)=R(x)-C(x)$ is the profit obtained from selling $x$ items, then the marginal profit $MP(x)$ is defined to be
$MP(x)=P^{\prime}(x)=MR(x)-MC(x)=R^{\prime}(x)-C^{\prime}(x)$ .

We can roughly approximate

M C (x) = C^{'} (x) = lim h \to 0 \frac{C (x + h) - C (x)}{h}

$MC(x)=C^{\prime}(x)=\underset{h\to 0}{\lim}\dfrac{C(x+h)-C(x)}{h}$

by choosing an appropriate value for $h$ .

Since $x$ represents objects, a reasonable and small value for $h$ is $1$ . Thus, by substituting $h=1$ , we get the approximation $MC(x)=C^{\prime}(x)\approx C(x+1)-C(x)$ .

Consequently, $C^{\prime}(x)$ for a given value of $x$ can be thought of as the change in cost associated with producing one additional item. In a similar way, $MR(x)=R^{\prime}(x)$ approximates the revenue obtained by selling one additional item, and $MP(x)=P^{\prime}(x)$ approximates the profit obtained by producing and selling one additional item.

Assume that the number of barbeque dinners that can be sold, $x$ , can be related to the price charged, $p$ , by the equation

$p(x)=9-0.03x, \, 0\le x\le 300$ .

In this case, the revenue in dollars obtained by selling $x$ barbeque dinners is given by

R (x) = x p (x) = x (9 - 0.03 x) = - 0.03 x^{2} + 9 x

$R(x)=xp(x)=x(9-0.03x)=-0.03x^2+9x$ for

0 \leq x \leq 300

$0\le x\le 300$ .

Use the marginal revenue function to estimate the revenue obtained from selling the $101$ ^st barbeque dinner.

Compare this to the actual revenue obtained from the sale of this dinner.

First, find the marginal revenue function: $MR(x)=R^{\prime}(x)=-0.06x+9$ .

Next, use $R^{\prime}(100)$ to approximate $R(101)-R(100)$ , the revenue obtained from the sale of the $101$ ^st dinner. Since $R^{\prime}(100)=3$ , the revenue obtained from the sale of the $101$ ^st dinner is approximately $$3$ .

The actual revenue obtained from the sale of the $101$ ^st dinner is

R (101) - R (100) = 602.97 - 600 = 2.97

$R(101)-R(100)=602.97-600=2.97$ , or

$ 2.97

$\$2.97$ .

The marginal revenue is a fairly good estimate in this case and has the advantage of being easy to compute.

Watch the following video to see the worked solution to this example.

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