MHB Constraint optimization without using Calculus

AI Thread Summary
The discussion revolves around maximizing the volume of a cylinder inscribed in a sphere with a radius of 6, using algebraic methods instead of calculus. The volume equation derived is V = 2πx(12 - x)(6 - x), indicating that the maximum occurs within the interval (0, 6). A suggestion is made to treat height (h) as the variable, leading to a new volume expression that can be maximized under the constraint from Pythagorean theorem. While the maximum volume can be calculated using calculus, alternatives like graphing calculators are proposed for a non-calculus approach. Ultimately, the conversation highlights the challenge of finding the maximum volume without calculus while exploring different methods.
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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 \(x = 2.57\).
 
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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.
 
Opalg said:
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.
 
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