# Optimization: Max Capacity of Cone-Shaped Cup

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• Nov 16th 2011, 09:12 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, 06:31 PM
skeeter
Re: Optimization: Max Capacity of Cone-Shaped Cup
$R$ = sector radius (a given constant)

$r$ = cone radius

$h$ = cone height

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

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

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

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

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

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

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

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

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