Chapter 4.8, Problem 47E

To determine

To calculate:The cost of producing 100 items. The cost of producing one item is $562 and the marginal cost of producing $x$ items is $1.92-0.002x$ .

Answer to Problem 47E

The cost of producing 100 items is $742.08.

Explanation of Solution

Given information:

The marginal cost of producing $x$ items is $1.92-0.002x$ and the cost of producing one item is $562.

Formula used:

The function $F$ is known as antiderivative of $f$ on an interval $I$ if $F^{\text{'}}\left(x\right)=f\left(x\right)$ for all $x$ in $I$ .

Calculation:

Consider the marginal cost of producing $x$ items are $1.92-0.002x$ and the marginal cost of

product is derivative of the cost of product.

Let cost of product is $C\left(x\right)$ then marginal cost is $C^{\text{'}}\left(x\right)=1.92-0.001x$ .

Integrate the equation on both sides,

  $\begin{array}{l}{\displaystyle \int C^{\text{'}}\left(x\right)dx}={\displaystyle \int 1.92-0.001x}dx\\ C\left(x\right)=1.92x-0.002\frac{x^2}{2}+D\end{array}$

Apply the condition that cost of one item is $562, $C\left(1\right)=\$562$ on the cost equation,

  $\begin{array}{c}C\left(1\right)=1.92\cdot 1-0.002\frac{1^2}{2}+D\\ 562=1.919+D\\ D=560.081\end{array}$

Put the value of $D=560.081$ in the cost equation,

  $C\left(x\right)=1.92x-0.002\frac{x^2}{2}+560.081$

So, the cost of producing $x=100$ items is

  $\begin{array}{c}C\left(100\right)=1.92\cdot 100-0.001{\left(100\right)}^2+560.081\\ \approx 742.08\end{array}$

Thus,the cost of producing 100 items is $742.08.