Chapter 4.4, Problem 36E

To determine

To find: To find the dimensions of cone.

Answer to Problem 36E

The dimensions of the rectangle radius $8$ cm, height $4$ cm and volume is $\frac{32\pi }{3}{\text{ cm}}^3$ .

Explanation of Solution

Given information : The cone is inscribed in a sphere of radius $3$ cm.

Formula used: $V=\frac{1}{3}(\pi )(r^2)(h)$

Let the radius of the base is r and h is the height.

  

The radius is $r^2=9-y^2$

The height is $h=y+3$

The volume of the cone is

  $\begin{array}{l}V=\frac{1}{3}(\pi )(r^2)(h)\\ V=\frac{1}{3}(\pi )(9-y^2)(y+3)\\ V=\frac{\pi }{3}(9y+27-y^3-3y^2)\end{array}$

For any function $f(x)$ the condition of minimum and maximum is obtained by finding $f\text{'}(x)=0$ , critical points and end points.

  $\begin{array}{l}\frac{dV}{dy}=\frac{d}{dy}\left(\frac{\pi }{3}(9y+27-y^3-3y^2)\right)\\ \frac{dV}{dy}=\frac{\pi }{3}(9-3y^2-6y)\end{array}$

Substitute $f\text{'}(x)=0$ in the above equation.

  $\begin{array}{l}0=\frac{\pi }{3}(9-3y^2-6y)\\ 0=\pi (3-y^2-2y)\\ 0=-\pi (y^2+2y-3)\\ 0=-\pi (y+3)(y-1)\end{array}$

Critical points occurs at

  $V\text{'}=0,-3<y<3$

  $\begin{array}{l}0=-\pi (y+3)(y-1)\\ y=1,y=-3\end{array}$

The critical point occurs at $y=1$ .

So, $\frac{dV}{dy}>0,-3<y<1$ and $\frac{dV}{dy}<0,1<y<3$ .

This corresponds to maximum volume.

For $y=1$

  $\begin{array}{l}h=y+3\\ h=1+3\\ h=4\end{array}$

The volume will be

  $\begin{array}{l}V=\frac{\pi }{3}(9y+27-y^3-3y^2)\\ V=\frac{\pi }{3}(9(1)+27-1^3-(3)1^2)\\ V=\frac{32\pi }{3}\end{array}$

Therefore, the radius will be

  $\begin{array}{l}r=9-y^2\\ r=9-1\\ r=8\end{array}$

For maximum volume, the dimensions are radius is $8$ cm, the height is $4$ cm and volume is $V=\frac{32\pi }{3}\text{ cm}$ .