Chapter 4, Problem 49RE

To determine

To find : The height and radius with smaller cone of the largest possible volume.

Answer to Problem 49RE

The answer:

  $\begin{array}{c}r=h\\ =4\end{array}$

Explanation of Solution

Given information:

The figure is:

  

Calculation:

  

Side view, The line formed by $(0,0)\text{ to }(6,12)$ is given by,

  $\begin{array}{c}y=\frac{12}{6}x\\ =2x,\\ \text{ or }x=\frac{1}{2}y\end{array}$

To describe the inner cone using the point $(x,y)$ of this line.

Then, $y=h$ and

  $\begin{array}{c}r=6-x\\ =6-\frac{1}{2}h\end{array}$

Obtain the cone volume equation as a function of a solitary variable

For $h$ , resolve the $r$ equation from above.

Enter the volume equation as a subtraction.

  $\begin{array}{c}r=6-\frac{1}{2}h\\ \Rightarrow 2r=12-h\\ \Rightarrow h=12-2r\\ V=\frac{1}{3}\pi r^{2}h\\ =\frac{1}{3}\pi r^2(12-2r)\\ =\frac{2}{3}\pi \left(6r^2-r^3\right)\end{array}$

To find $V^{\prime }$ where it is $=0$

  $\begin{array}{c}{V}^{\prime }=\frac{2}{3}\pi \left(12r-3r^2\right)\\ =2\pi r(4-r)\\ \Rightarrow 0=2\pi r(4-r)\\ \Rightarrow r=0,4\end{array}$

Plug it back into the equation relating $r$ and $h$ above to find $h$ :

  $\begin{array}{c}h=12-2r\\ =12-2(4)\\ =4\end{array}$

Therefore, the required height and radius with smaller cone of the largest possible volume:

  $\begin{array}{c}r=h\\ =4\end{array}$