Chapter 5.4, Problem 32E

To determine

To find the radius, height and volume of the cone.

Answer to Problem 32E

The volume of the radius $\sqrt{2}$ , height $1$ , and volume of the cone is $\frac{2\pi }{3}$ .

Explanation of Solution

Given information: Given volume of a cone is $V=\frac{1}{3}\pi r^{2}h.$

Calculation:

The volume of a cone is $V=\frac{1}{3}\pi r^{2}h.$ before find the maximum of this equation, eliminate one of the variables. Using the right triangle and Pythagorean Theorem,

  $\begin{array}{l}{h}^2+r^2={\left(\sqrt{3}\right)}^2\\ r^2=3-h^2\end{array}$

Substitute in $r^2=3-h^2$ into the equation for the volume then givens;

  $\begin{array}{l}V=\frac{1}{3}\pi \left(3-h^2\right)h\\ =\frac{1}{3}\left(3\pi h-\pi h^3\right)\\ =\pi h-\frac{1}{3}\pi h^3\\ v=\frac{\pi r^{2}h}{3}\\ =\pi \frac{\left(3-h^2\right)h}{3}\end{array}$

Then max the function $f\left(x\right)=x\left(3-x^2\right)$ for positive $x<\sqrt{3}$

  $\begin{array}{l}{f}^{\prime }=3-x^2-2x^2\\ =3-3x^2\end{array}$

Only one solution acceptable $x=1$ ,the $h=1,r=\sqrt{3-1}\text{ = }\sqrt{2}$ and $V=\frac{2\pi }{3}$

Thus, the volume of the radius $\sqrt{2}$ , height $1$ , and volume of the cone is $\frac{2\pi }{3}$ .