Chapter 2, Problem 91RE

Solution

(a)

To find: The volume V as the function of radius x.

The volume V as the function of radius x.is $V=140\pi x^2-\frac{2\pi x^3}{3}$

Given information

The total length of the container is 140 ft.The radius of semi hemisphere is x.

  

In the above figure the diameter of the hemisphere is 2x and radius is x.The volume of hemisphere is $\frac{2}{3}\pi x^3$ .There are two hemisphere on both end.Hence total volume of two hemisphere is $\frac{4}{3}\pi x^3$

Volume of cylinder whose height is 140-2x and radius is x

  $\pi x^2\left(140-2x\right)$

Hence total volume of the container is :

  $\begin{array}{l}V=\frac{4}{3}\pi x^3+\pi x^2\left(140-2x\right)\\ V=\frac{4}{3}\pi x^3-2\pi x^3+140\pi x^2\\ V=140\pi x^2-\frac{2\pi x^3}{3}\end{array}$

(b)

To find: graph of function $y=V\left(x\right)$

The graph is shown.

Given information The volume V as the function of radius x.is $V=140\pi x^2-\frac{2\pi x^3}{3}$

  

(c)

To find: The radius of container with the largest possible volume.

The volume V is largest at x=140 ft.

Given information The total length of the container is 140 ft.The radius of semi hemisphere is x.the volume of the container is $V=140\pi x^2-\frac{2\pi x^3}{3}$

Differentiate V with respect to x;

  $\frac{dV}{dx}=280\pi x-\frac{6\pi x^2}{3}$

Now calculate $\frac{dV}{dx}=0$

  $\begin{array}{l}280\pi x-\frac{6\pi x^2}{3}=0\\ 280\pi x=\frac{6\pi x^2}{3}\\ 280=2x\\ x=140\end{array}$

Now find $\frac{d^{2}V}{d{x}^2}$

  $\frac{d^{2}V}{d{x}^2}=280\pi -4\pi x$

Substitute x=140

  $\begin{array}{l}{\left(\frac{d^{2}V}{d{x}^2}\right)}_{x=140}=280\pi -4\pi \left(140\right)\\ {\left(\frac{d^{2}V}{d{x}^2}\right)}_{x=140}=280\pi -560\pi \\ {\left(\frac{d^{2}V}{d{x}^2}\right)}_{x=140}=-280\pi <0\end{array}$

Hence the volume is maximum at x=140

The radius of the container is 140 ft.