Chapter 5.4, Problem 26E

To determine

Find the production level that minimizes average cost.

Answer to Problem 26E

The production level $x=5000$ average cost is minimum.

Explanation of Solution

Given information: Given the cost is $c\left(x\right)=x{e}^x-2x^2$ .

Calculation:

The average cost is,

  $\begin{array}{l}{c}_{avg}=\frac{c\left(x\right)}{x}\\ =\frac{x{e}^x-2x^2}{x}\\ =e^x-2x\end{array}$

The average cost will be minimum or maximum at ${c^{\prime }}_{avg}\left(x\right)=0$ and the average cost will be minimum if ${c^{\prime }}_{avg}"\left(x\right)>0$

  $\begin{array}{l}{c}_{avg}\left(x\right)=e^x-2x\\ {c^{\prime }}_{avg}\left(x\right)=0\\ e^x-2=0\\ x=\ln 2\\ =0.693\\ {c^{\prime }}_{avg}\left(x\right)=e^x-2\\ c_{avg}"\left(x\right)=e^x>0\end{array}$

It shows that $x=0.693$ corresponds to minimum average cost. Since $x$ is measured in thousands of units $x=693$ .

Thus, the production level $x=693$ average cost is minimum