Answer to Question #280664 in Calculus for Ash

Question

For the following exercises, draw the given optimization problem and solve.

1) Find the volume of the largest right circular cylinder that fits in a sphere of radius 1.

Answer 1

Let $R=$ the radius of the sphere, $r=$ the radius of a base of a inscribed cylinder, $h=$ the height of a cylinder.

Then $V_{cyl}=\pi r^2 h.$

By the Pythagorean Theorem

$$R^2=r^2+(h/2)^2\ \ const$$

$$r^2=R^2-(h/2)^2$$

Substitute

$$V_{cyl}=V(h)=\pi (R^2-(h/2)^2) h$$

$$V(h)=\dfrac{\pi}{4}(4R^2h-h^3) , 0<h<2R$$

Find the first derivative with respect to $h$

$$V'(h)=\dfrac{\pi}{4}(4R^2-3h^2)$$

Find the critical number(s)

$$V'(h)=0=>\dfrac{\pi}{4}(4R^2-3h^2)=0$$

$$h_1=-\dfrac{2}{\sqrt{3}}R, h_2=\dfrac{2}{\sqrt{3}}R$$

If $0<h<\dfrac{2}{\sqrt{3}}R, V'(h)>0, V(h)$ increases.

If $\dfrac{2}{\sqrt{3}}R<h<2R, V'(h)<0, V(h)$ decreases.

The function $V(h)$ has the absolute maximum on $[0, 2R]$ at $h=\dfrac{2}{\sqrt{3}}R.$

$$r^2=R^2-(\dfrac{2}{2\sqrt{3}}R)^2=\dfrac{2}{3}R^2$$

$$r=\dfrac{\sqrt{2}}{\sqrt{3}}R$$

$$V_{cyl\ max}=\pi(\dfrac{2}{3}R^2)(\dfrac{2}{\sqrt{3}}R)$$

$$=\dfrac{4\sqrt{3}\pi}{9}R^3({units}^3)$$

If $R=1,$ then $V_{cyl\ max}=\dfrac{4\sqrt{3}\pi}{9}{units}^3.$