Finding a maximal cylinder to be placed in a sphere

Question

Take both volumes, viz.: pi r^2 h to be < 4/3 pi a^3, to distinguish between radii of the 2 figures, let us call the radius of the sphere a

pi is divided out

r^2 h < 4/3 a^3

It is time to take the derivate in order to find the maximal r, but the 4/3 a^3 in case of differentiating on r, is a constant, and is lost in the derivative. With thus all useful info is lost. But I see no way to retain the constant in the derivative. Or is there a cleverer way? thank you

Answer 1

What you're missing is you haven't actually made any link between r,h and a.

I presume a is constant here, and you want the r and h of a cylinder that (a) fits inside the sphere, and (b) has maximal volume. Yes?

Intuitively, the top and bottom of the cylinder will be jammed against the internal walls of the sphere. If you draw a side-on diagram, you'll see a right-angle triangle with sides $r$ and $h/2$, and hypotenuse $a$. So, you get $a^2=r^2 + h^2/4$.

Now, you can either

• rearrange that formula, substitute it into the equation for the cylinder's volume to get a formula involving either $r$ only or $h$ only, then differentiate that to find the $r$ or $h$ that maximises the volume
• differentiate both $V=\pi r^2 h$ and $a^2=r^2+h^2/4$ implicitly, to get two equations for $dV/dh$ and $dr/dh$ (or $dV/dr$ and $dh/dr$). Then set $dV/dh$ (or $dV/dr$) to zero, eliminate the other derivative and one of $h$ and $r$ (using $a^2=r^2+h^2/4$), and you'll have the other of $h$ and $r$.