# Optimization: Max Capacity of Cone-Shaped Cup

• Nov 16th 2011, 08:12 AM
AXQ4286
Optimization: Max Capacity of Cone-Shaped Cup
A cone shaped drinking cup is made from a circular piece of paper of radius R by cutting out a sector and joining the edges CA and CB. Find the maximum capacity of such a cup.
• Nov 16th 2011, 05:31 PM
skeeter
Re: Optimization: Max Capacity of Cone-Shaped Cup
$\displaystyle R$ = sector radius (a given constant)

$\displaystyle r$ = cone radius

$\displaystyle h$ = cone height

$\displaystyle R^2 = r^2 + h^2$

$\displaystyle r^2 = R^2 - h^2$

$\displaystyle V = \frac{\pi}{3} r^2 h$

$\displaystyle V = \frac{\pi}{3} (R^2 - h^2)h$

$\displaystyle V = \frac{\pi}{3} (R^2 h - h^3)$

$\displaystyle \frac{dV}{dh} = \frac{\pi}{3} (R^2 - 3h^2)$

$\displaystyle \frac{dV}{dh} = 0$ and is a maximum when $\displaystyle h = \frac{R}{\sqrt{3}}$ since $\displaystyle \frac{d^2V}{dh^2} < 0$ for all $\displaystyle h$

$\displaystyle V_{max} = \frac{\pi}{3} \left(R^2 - \frac{R^2}{3}\right) \frac{R}{\sqrt{3}}$

$\displaystyle V_{max} = \frac{2 \pi R^3}{9\sqrt{3}}$