Chapter 4.1, Problem 43E

(a)

To determine

To find : extreme values of $V$ .

Answer to Problem 43E

Maximum occur at $\left(2,144\right)$ .

Explanation of Solution

Given information :

  $V\left(x\right)=x\left(10-2x\right)\left(16-2x\right),0<x<5$

Formula needed :

For critical points, $\frac{dy}{dx}=0$ or $\frac{dy}{dx}$ is not defined.

Simplify $V\left(x\right)=x\left(10-2x\right)\left(16-2x\right)$ .

  $\begin{array}{c}V\left(x\right)=\left(10x-2x^2\right)\left(16-2x\right)\\ =10x\left(16-2x\right)-2x^2\left(16-2x\right)\\ =160x-20x^2-32x^2+4x^3\\ =160x-52x^2+4x^3\end{array}$

Differentiate $V\left(x\right)=160x-52x^2+4x^3$ with respect to $x$ .

  $\begin{array}{c}{V}^{\prime }\left(x\right)=\frac{d}{dx}\left(160x-52x^2+4x^3\right)\\ =160-104x+12x^2\end{array}$

Take $V^{\prime }\left(x\right)=0$ for critical points.

  $\begin{array}{c}160-104x+12x^2=0\\ 4\left(40-26x+3x^2\right)=0\left[4\ne 0\right]\\ 40-26x+3x^2=0\\ 40-20x-6x+3x^2=0\end{array}$

Further simplify,

  $\begin{array}{c}20\left(2-x\right)-3x\left(2-x\right)=0\\ \left(20-3x\right)\left(2-x\right)=0\\ x=\frac{20}{3},2\end{array}$

Only one critical point exists in the given interval.

Plug the critical point in the function.

  $\begin{array}{c}V\left(2\right)=160\left(2\right)-52{\left(2\right)}^2+4{\left(2\right)}^3\\ =320-52\cdot 4+4\cdot 8\\ =320-208+32\\ =112+32\\ =144\end{array}$

Hence, the function has maximum at $\left(2,144\right)$ .

(b)

To determine

To interpret : any values found in part (a).

Explanation of Solution

Given information :

  $V\left(x\right)=x\left(10-2x\right)\left(16-2x\right),0<x<5$

Formula needed :

For critical points, $\frac{dy}{dx}=0$ or $\frac{dy}{dx}$ is not defined.

The function has maximum at $\left(2,144\right)$ . This means the largest volume of box possible in the given interval is $144$ when $x=2$ .

(c)

To determine

To graph : any values found in part (a) and part (b).

Explanation of Solution

Given information :

  $V\left(x\right)=x\left(10-2x\right)\left(16-2x\right),0<x<5$

Formula needed :

For critical points, $\frac{dy}{dx}=0$ or $\frac{dy}{dx}$ is not defined.

Let us draw the graph of function and derivative of function on same in the interval $0<x<5$ .

  

The curve in green is denoting function and curve in red is denoting derivative of function.

The largest volume of box possible at $x=2$ .