I have given to compute the maximum volume of a cone inscribed in a sphere of radius $R>0$.
My question is give that x-radius of the cone the formula to maximize is
$\frac{1}{3} \pi (R^2-x^2)(R+x)$ or $\frac{1}{3} \pi x^2 \times \sqrt{R^2-x^2}$?
[Math] Max volume of a cone in a sphere
calculusderivativesoptimization
Best Answer
If the radius of the cone is $x\leq R$ then the distance of the base of the cone from the center of the sphere is given by $\sqrt{R^2-x^2}$ and the height of the cone is $R+\sqrt{R^2-x^2}$. If we set $x=R\sin\theta$, we have to maximize:
$$ V = \frac{\pi}{3} x^2\left(R+\sqrt{R^2-x^2}\right) = \frac{\pi R^3}{3}\sin^2\theta\,(1+\cos\theta) $$ or, by setting $y=\cos\theta$, $$ V = \frac{\pi R^3}{3}(1-y^2)(1+y). $$ Since $\frac{d}{dy}(1+y)(1-y^2)=(1+y)(1-3y)$, the maximum is achieved at $y=\frac{1}{3}$ and it is: $$ V = \frac{32}{81}\pi R^3.$$