Constraint optimization without using Calculus

[Dustinsfl]

A student in hs I tutor was giving the following problem:

Maximizes the volume of a cylinder inscribe in a sphere of radius 6.

We worked through it and had:
\begin{align}
h &= 2(6 - x)\\
r_{\text{cylinder}}^2 &= x(12 - x)
\end{align}
Now the volume of a right cylinder is \(V = \pi r^2h\) so
\[
V = 2\pi x(12 - x)(6 - x).
\]
Since this is a cubic of the form \(x^3\) and not \(-x^3\), we know that the maximum occurs when \(x\in(0, 6)\).

How are we supposed to find this value of x without Calculus? I ended up taking the derivative to determine it and it was something like \(x = 2.57\).

 

[Opalg]

I would take $h$ as the variable rather than $x$. You want to maximise $V = \pi r^2h$ subject to the condition $r^2 + \bigl(\frac h2\bigr)^2 = 6^2$, as you can see from Pythagoras' theorem in the circle formed by a cross-section of the sphere through its centre. So you want to maximise $\pi h\Bigl(36 - \dfrac{h^2}4\Bigr)$. The derivative vanishes when $h = 4\sqrt3$, giving $V = 48\sqrt3\pi$. But I don't see how that could be done without calculus.

 

[Prove It]

I would take $h$ as the variable rather than $x$. You want to maximise $V = \pi r^2h$ subject to the condition $r^2 + \bigl(\frac h2\bigr)^2 = 6^2$, as you can see from Pythagoras' theorem in the circle formed by a cross-section of the sphere through its centre. So you want to maximise $\pi h\Bigl(36 - \dfrac{h^2}4\Bigr)$. The derivative vanishes when $h = 4\sqrt3$, giving $V = 48\sqrt3\pi$. But I don't see how that could be done without calculus.

You could always do a plot using a graphing calculator and read off the point of the maximum. It's not an exact method, but it's probably enough if it's to be done without calculus.