Chapter 4.4, Problem 24E

To determine

To find: Whether the production level maximizes the average cost or not and hence it’s value.

Answer to Problem 24E

Yes, the production level maximizes the average cost at $x=1.525$ .

Explanation of Solution

Given information:

  $r\left(x\right)=\frac{x^2}{x^2+1}$ , where $r\left(x\right)$ is a revenue function.

  $c\left(x\right)=\frac{{\left(x-1\right)}^3}{3}-\frac{1}{3}$ , where $c\left(x\right)$ is a revenue function.

Concept used: The maximum value of the function $f\left(x\right)$ will be at the critical point of that function.

Calculation:

Subtract the cost function from the revenue function to obtain the profit function, $p\left(x\right)$ .

  $\begin{array}{c}p\left(x\right)=r\left(x\right)-c\left(x\right)\\ =\frac{x^2}{x^2+1}-\left(\frac{{\left(x-1\right)}^3}{3}+\frac{1}{3}\right)\end{array}$

Differentiate the obtained function with respect to $x$ .

  $\begin{array}{c}{p}^{\prime }\left(x\right)=\frac{d}{dx}\left(\frac{x^2}{x^2+1}\right)-\frac{d}{dx}\left(\frac{{\left(x-1\right)}^3}{3}+\frac{1}{3}\right)\\ =\frac{2x}{{\left(x^2+1\right)}^2}-{\left(x-1\right)}^2\end{array}$

Equate the obtained derivative to 0 and solve for $x$ .

  $\begin{array}{c}{p}^{\prime }\left(x\right)=0\\ \frac{2x}{{\left(x^2+1\right)}^2}-{\left(x-1\right)}^2=0\\ x\approx 0.294,1.525\end{array}$

Now, differentiate the equation, $p^{\prime }\left(x\right)=\frac{2x}{{\left(x^2+1\right)}^2}-{\left(x-1\right)}^2$ , with respect to $x$ .

  $p^{″}\left(x\right)=\frac{2\left(-3x^2+1\right)}{{\left(x^2+1\right)}^3}-2\left(x-1\right)$

Substitute $x=0.294$ and $x=1.525$ , separately into the obtained second derivative.

  $\begin{array}{c}{p}^{″}\left(0.294\right)=\frac{2\left(-3{\left(0.294\right)}^2+1\right)}{{\left({\left(0.294\right)}^2+1\right)}^3}-2\left(\left(0.294\right)-1\right)\\ \approx 2.6\end{array}$

  $\begin{array}{c}{p}^{″}\left(1.525\right)=\frac{2\left(-3{\left(1.525\right)}^2+1\right)}{{\left({\left(1.525\right)}^2+1\right)}^3}-2\left(\left(1.525\right)-1\right)\\ \approx -\text{1}\text{.375}\end{array}$

Since at $x=1.525$ , $p^{″}\left(x\right)<0$ , the profit will be maximum at $x=1.525$ .

So, at $x=1.525$ , the production level will be maximum.