Question

Find the volume of the largest right circular cone that can be inscribed in a sphere of radius 3?

Answer 1

The answer is 32 pie over 3

Answer 2

First, notice that, by the Pythagorean Theorem,

$x^2+y^2=3^2$

meaning that:

$x^2=9-y^2$

Also, since the volume of a cone with radius r and height h is $\frac{1}{3} \pi r^2h$ we know that the volume of the cone is:

$\frac{1}{3} \pi x^2 (3+y) = \frac{1}{3} \pi (9-y^2)(3+y) = \frac{1}{3} \pi [27+9y-3y^2-y^3]$

Therefore, we want to maximize the function $V(y) = \frac{1}{3} \pi [27+9y-3y^2-y^3]$ subject to the constraint $0 \leq y \leq 3$.

To find the critical points, we differentiate:

$V'(y)= \frac{1}{3} \pi [9-6y-3y^2] = \pi [3-2y-y^2] = \pi (3+y)(1-y).$

Therefore, $V'(y) = 0$ when

                                                    $\pi (3+y)(1-y)=0$

meaning that $y = -3$ or $y=1$. Only $y=1$ is in the interval $[0,3]$ so that’s the only critical point we need to concern ourselves with.

Now we evaluate $V$ at the critical point and the endpoints:

$V(0) = \frac{1}{3} \pi [27+9(0) - 3(0)^2] = 9 \pi$

$V(1) = \frac{1}{3} \pi [27+9(1)-3(1)^2-1^3] = \frac{32 \pi }{3}$

$V(3) = \frac{1}{3} \pi [27+9(3) - 3(3)^2-3^2] = 0$

Therefore, the volume of the largest cone that can be inscribed in a sphere of radius 3 is $\frac{32 \pi }{3}$