Chapter 3.3, Problem 91AYU

Minimizing Marginal Cost The marginal cost of a product can be thought of as the cost of producing one additional unit of output. For example, if the marginal cost of producing the 50th product is $\$6.20$, it cost $\$6.20$ to increase production from 49 to 50 units of output. Suppose the marginal cost $C$ (in dollars) to produce $x$ thousand digital music players is given by the function

$C(x)=x^2-140x+7400$

(a) How many players should be produced to minimize the marginal cost?

(b) What is the minimum marginal cost?

To determine

To calculate: The minimum marginal cost and the number of units to be produced so as to reduce the marginal cost.

Answer to Problem 91AYU

The 70,000 digital music players (since $x$ is in thousands) have to be produced in order to minimize the marginal cost.

The minimum marginal cost will be $\$2500$.

Explanation of Solution

Given:

The marginal cost $C$ to produce $x$ thousand digital music players is given by the function

$C(x) = x^2 - 140x + 7400$

Formula used:

Consider a quadratic function of the form $f(x) = a{x}^2 + bx + c, a \ne 0$.

Then the graph of the above function is a parabola with vertex $\left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right)$.

This vertex is the highest point if $a < 0$ and the lowest point if $a > 0$.

Therefore, the maximum (or the minimum) value of the function will be $f\left(\frac{-b}{2a}\right)$.

Calculation:

The given function$C(x) = x^2 - 140x + 7400$ is a quadratic function.

Thus, here, we have $a = 1 > 0$.

Therefore, the vertex is the minimum point in the graph of the given function.

Thus, we have

$\frac{-b}{2a} = \frac{-(-140)}{2(1)} = 70$.

$C\left(\frac{-b}{2a}\right) = C(70) = {70}^2 - 140(70) + 7400 = 2500$.

Thus, we get the vertex of the function as$(70, 2500)$.

Therefore,

c.The 70,000 digital music players (since $x$ is in thousands) have to be produced in order to minimize the marginal cost.

d.The minimum marginal cost will be $\$2500$.