Cylinder In A Sphere

Relevant wiki: Applying the Arithmetic Mean Geometric Mean Inequality

I see that Rocco has posted my favorite approach (calculus). Let me post an classical inequalities approach.

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Let $r$ and $h$ denote the radius and height of the inscribed cylinder.
And let $R$ denote the radius of the sphere of unit volume.

By recalling the formula for the volume of a cylinder and the volume of a sphere ,
we have $V_{\text{cylinder}} = \pi r^2 h$ and $V_{\text{sphere}} = \dfrac43 \pi R^3 = 1$ , respectively.

First solving for $R$ gives $R = \sqrt[3]{\dfrac3{4\pi} }$ .

And we want to evaluate $\max(\pi r^2 h) = \pi \max(r^2 h)$ .

By Pythagorean theorem , $(2r)^2 + h^2 =(2R)^2$ or equivalently $4r^2 + h^2 = 4\left( \sqrt[3]{\dfrac3{4\pi} }\right)^2$ .

By applying the weighted AM-GM inequality on the numbers $2r^2, 2r^2 , h^2$ , we have

$AM (2r^2 ,2r^2 , h^2 ) \geq GM(2r^2, 2r^2 ,h^2) \qquad \Rightarrow \qquad \dfrac{2r^2 + 2r^2 + h^2}{3} \geq \sqrt[3]{(2r^2)^2 (h^2)} = \sqrt[3]{4} (r^2 h)^{2/3}$

Upon simplifying,

$\dfrac43 \left( \sqrt[3]{\dfrac3{4\pi} }\right)^2 \cdot \dfrac1{\sqrt[3]{4}} \geq (r^2 h)^{2/3} \qquad \Rightarrow \qquad r^2 h \leq \dfrac1{\sqrt3 \pi} .$

Hence $\pi \max(r^2 h) = \pi \cdot \dfrac1{\sqrt3 \pi} = \dfrac1{\sqrt3}$ .

The equality holds when $2r^2 = h^2 \Leftrightarrow \dfrac hr = \sqrt2 \Leftrightarrow r = \dfrac1{\sqrt[6]{6} \sqrt[3]{\pi} } , h = \dfrac{\sqrt[3]{\frac2\pi}}{\sqrt[6]{3}}$ .

Our answer is $a+b= 1+3 = \boxed4$ .

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Food for thought : How did I choose the numbers $2r^2, 2r^2, h^2$ in the weighted AM-GM inequality?

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EDIT : Oh! After posting my solution, I discovered that it's much simpler to solve this problem by using Lagrange multipliers . Anyone wants to give it a try?

Let the volume of the sphere ${\bf V_s = \frac{4}{3}\pi r^3 }$ $$ and the volume of the cylinder ${\bf V_c = \pi R^2H }$

$$ From the geometry of the problem ${\bf r^2 = (\frac{H}{2})^{2} + R^2 \implies R^2 = \frac{4r^2 - H^2}{4} }$

${\bf \implies R = \frac{\sqrt{4r^2 - H^2}}{2} \implies V_c = \frac{\pi}{4}(4r^2H - H^3) \implies }$

${\bf \frac{dV_c}{dH} = \frac{\pi}{4}(4r^2 - 3H^2) = 0 \implies H = \frac{2}{\sqrt{3}}r }$

${\bf \frac{d^2V_c}{dH^2}|_(H = \frac{2}{\sqrt{3}}r) = -\sqrt{3}\pi r < 0 \implies }$ max occurs at ${\bf H = \frac{2}{\sqrt{3}} r. }$

${\bf H = \frac{2}{\sqrt{3}} r \implies R^2 = \frac{2}{3}r^2 \implies R = \sqrt{\frac{2}{3}}r \implies }$

${\bf V_c(max) = \frac{4}{3 * \sqrt{3}}\pi r^3 = \frac{1}{\sqrt{3}}V_s. }$ $$

Since ${\bf V_s = 1 \implies V_c(max) = \frac{1}{\sqrt{3}} = \frac{a}{\sqrt{b}} \implies a + b = 4. }$