# optimization

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• Nov 12th 2008, 10:50 AM
thecount
optimization
a cone shaped drinking cup id made from a circular piece of paper of radius R by cutting a sector and joining the edges CA and CB. fond the maximum capacity for such a cup.

i know the answer is V=2 pi R^2/ (9 sqrt 3)

but i dont know how to get there...
• Nov 12th 2008, 11:47 AM
earboth
Quote:

Originally Posted by thecount
a cone shaped drinking cup id made from a circular piece of paper of radius R by cutting a sector and joining the edges CA and CB. fond the maximum capacity for such a cup.

i know the answer is V=2 pi R^2/ (9 sqrt 3)

but i dont know how to get there...

1. Your answer can't be right because a volume has the dimension 3 but your term at the RHS has the dimension 2.

2. The volume of the cone is calculated by:

$V=\frac13 \pi r^2\cdot h$

3. The radius of the piece of paper, the radius of the cone and the height of the cone form a right triangle:

$r^2+h^2=R^2~\implies~r^2=R^2-h^2$

4. Substitute into the equation of the volume. You'll get a function wrt h:

$V(h)=\frac13 \pi (R^2-h^2)\cdot h=-\frac13\pi h^3+\frac13\pi R^2\cdot h$

5. Solve for h: V'(h) = 0. I've got $h = \frac13R\sqrt{3}$

6. Re-substitute to calculate r. I've got $r=\frac13R\sqrt{6}$

7. Plug in these values to calculate the maximum volume:

$V=\frac13 \pi r^2\cdot h~\implies~V=\frac13 \pi \frac23 R^2\cdot \frac13 R \sqrt{3} = \frac2{27} \pi R^3 \sqrt{3}$