I have a question that I have no idea how to proceed in. Could use some insight:

A right circular cylinder is inscribed in a sphere of radius r. Find the largest possible volume of such a cylinder.

My hint is to:

Draw the appropriate right triangle and the Pythagorean Theorem will connect all of the variables.

So the Area of a cylinder is $\pi \cdot r^2 \cdot h$. What next?

Does the right triangle have the sides of $h$, $r$, and then $\sqrt{h^2 + r^2}$ as the hypotenuse? But where does that get us in terms of finding the Volume of the cylinder in one variable and then taking the derivative?

  • $\begingroup$ The sphere has radius $r$. The radius of the inscribed cylinder must be smaller. Drawing a sketch might help. $\endgroup$ Mar 10, 2018 at 16:27

2 Answers 2


More than a hint...If $R$ is the radius of the sphere and $r$ is the radius of the cylinder, with $h$ the height of the cylinder, then by Pythagoras we have $$\frac{h^2}{4}=R^2-r^2$$

The volume of the cylinder is then $$V=2\pi r^2\sqrt{R^2-r^2}$$ $$\implies V^2=4\pi^2(r^4R^2-r^6)$$

You can differentiate this to get $$2VV'=4\pi^2(4r^3R^2-6r^5)=0$$ So the maximum $V$ which is when $r=R\sqrt{\frac 23},$ and $$V_{max}=\frac{4\pi R^3}{3\sqrt{3}}$$

  • $\begingroup$ I cant see how you got the first equation by Pythagoras $\endgroup$
    – Jwan622
    Mar 10, 2018 at 22:57
  • $\begingroup$ Have you tried drawing a picture? $\endgroup$ Mar 10, 2018 at 22:59
  • $\begingroup$ I have....I have that the diameter of a sphere is $\sqrt{h^2 + 4r^2}$ $\endgroup$
    – Jwan622
    Mar 10, 2018 at 23:02
  • $\begingroup$ Oh I see... you're doing $h^2 + (2r)^2 = (2R)^2$ $\endgroup$
    – Jwan622
    Mar 10, 2018 at 23:09
  • $\begingroup$ So that's the same equation as I wrote since the diameter of the sphere is $2R$ $\endgroup$ Mar 10, 2018 at 23:09

Giving outline of Lagrange multiplier method, quick and threadbare

constraint dia

$$ D=\sqrt{4r^2+h^2}$$

To maximize volume

$$V= \pi r^2 h \implies r^2 h $$ only need be considered without multiplicative const.

$$ \frac{D_r}{D_h}=\frac{V_r}{V_h}$$

$$ \frac{2rh}{r^2}=\frac{8r}{2h}\rightarrow h=\sqrt2 r$$

fixes relative proportions.To link to sphere radius:

$$ D= 2R = \sqrt{4r^2+2r^2}\rightarrow \frac{r}{R}= \sqrt{\frac{2}{3}}$$


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