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 \(\frac{32 \pi }{3}\)
