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

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 $$\frac{32 \pi }{3}$$