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

We can roughly approximate

[latex]MC(x)=C^{\prime}(x)=\underset{h\to 0}{\lim}\dfrac{C(x+h)-C(x)}{h}[/latex]

by choosing an appropriate value for [latex]h[/latex].

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

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

Assume that the number of barbeque dinners that can be sold, [latex]x[/latex], can be related to the price charged, [latex]p[/latex], by the equation

[latex]p(x)=9-0.03x, \, 0\le x\le 300[/latex].

In this case, the revenue in dollars obtained by selling [latex]x[/latex] barbeque dinners is given by

[latex]R(x)=xp(x)=x(9-0.03x)=-0.03x^2+9x[/latex] for [latex]0\le x\le 300[/latex].

Use the marginal revenue function to estimate the revenue obtained from selling the [latex]101[/latex]st barbeque dinner.

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