A cylindrical frame volume max

• Dec 14th 2006, 08:51 PM
Sky12
A cylindrical frame volume max
A cylindrical kite frame is to be constructed from a 4-m length of light bendable rod.

The frame will be made up of 2 circles joined by four straight rods of equal length. In order to maximize lift, the kite frame must be constructed to maximize the volume of the cylinder.

In what lengths should the peices be cut to optimize the kites flight?

I know the answer but every time I try to do it, I get a different one, help would be appreicated greatly.
• Dec 14th 2006, 10:24 PM
earboth
Quote:

Originally Posted by Sky12
A cylindrical kite frame is to be constructed from a 4-m length of light bendable rod.

The frame will be made up of 2 circles joined by four straight rods of equal length. In order to maximize lift, the kite frame must be constructed to maximize the volume of the cylinder.

In what lengths should the peices be cut to optimize the kites flight?

I know the answer but every time I try to do it, I get a different one, help would be appreicated greatly.

Hello,

The volume of a cylindre is calculated by:

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

You've got the additional condition:

$4 = \underbrace{2 \cdot 2 \pi r}_{\text{two circles}}+4h$
Solve this equation for h:

$h=1-\pi r$. Now plug in this term for h into the first equation and expand:

$V=\pi \cdot r^2 \cdot (1-\pi r)=\pi r^2- \pi^2 \cdot r^3$

The volume has an extreme value if the first derivative equals zero:

$V'(r)=2 \pi r-3 \pi^2 r^2=\pi r(2-3\pi r)$

Now solve the equation V'(r) = 0 for r. You'll get r = 0. This kite has the volume zero, which is indeed an extremum. The 2nd value for r is:
$r=\frac{2}{3 \pi}\ \Longrightarrow \ h=\frac{1}{3}$

The maximal volume is $V_{max}=0.047 \text{ m³}=12.4\text{ gal(US)}$

EB