I am working through the following problem:
In $R^3$, consider a sphere of radius r. If one makes a longitudinal cut (i.e., perpendicular to a radius) at a distance r − h from the center, the cut-off portion has height h. Given a fixed surface area A > 0 of the cut-off portion, find r and h to maximize the volume of the cut-off pieces.
This yields the nonlinear optimization problem: \begin{align} \min_{r,h \in R} \quad & -\pi h^2 (r-h/3) \\ st \quad & \ 2 \pi rh = A \\ & \ r, h \ge 0. \end{align}
I found the Lagrangian: $$ L(r,h,\mu_1) = -\pi h^2 (r - \frac{h}{3}) + \mu_1(2\pi r h-A)$$ and KKT conditions $$\frac{\partial L}{\partial r} = -\pi h^2 + 2\mu_1\pi h = -h^2 + 2\mu_1 h=0 $$ $$\frac{\partial L}{\partial h} = -2\pi r h + \pi h^2 + 2\mu_1\pi r = -2rh+h^2-2r\mu_1=0 $$ $$2\pi r h =A$$ $$\mu_1 \ge 0$$
Then I found a KKT point at $\mu_1 = \frac{1}{2}, h=1, r=1$. However, I couldnt find any more KKT points. The problem asks for an optimal solution. The problem is not convex optimization, so we can not say the KKT point is globally optimal. I tried to model the problem in Gurobi but couldn't linearize the objective function. How can I find the optimal solution to this nonlinear problem?
Hint:
The simplest way of solving this problem is to use the constraint $\ 2\pi rh=A\ $ to reduce the dimension of the objective function's domain. Substituting $\ r= \frac{A}{2\pi h}\ $ into the function $\ \pi h^2\left(r-\frac{h}{3}\right)\ $ reduces it to $ \frac{Ah}{2}-\frac{\pi h^3}{3}\ $. There is a unique non-negative point at which the derivative of this function vanishes, and you should be able to show that this point is a maximum by investigating the sign of the derivative.