Optimization: Cylinderical Kite Frame

[wind]

Hi, I really need help with this question

A cylindrical kite frame is to be constructed from a 4m lenght of light bendable rod. the frame will be made up of two circles joined by four straight rods of equal lenght. in order to maximize lift, the kite frame must be constructed to maximize the volume of the cylinder. Into what lenghts should the pieces be cut in order to optimize the kite's flight?

Total Perimiter

4=4h+4pir
4/4pir=4h
4/16pir=h
1/4pir=h

Maximize Volume

V=pir²h
V=pir²(1/4pir)
V=pir²/4pir
V=r/4

help

 

[skeeter]

let x = length of rod material used to make the two circular ends

4 - x = length of rod material used to make the four straight rods

x/2 = 2pi*r = circumference of one circular end, so r = x/(4pi)

area of a circular end = pi*r² = x²/(16pi)

height of the cylinder = (4 - x)/4

V = x²/(16pi)*(4 - x)/4 = x²(4 - x)/(64pi)

find dV/dx and maximize.

 

[soroban]

Hello, wind!

Your intentions were correct . . . a little algebra was off.

A cylindrical kite frame is to be constructed from a 4m lenght of light bendable rod.
The frame will be made up of two 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.
Into what lengths should the pieces be cut in order to optimize the kite's flight?

Total Perimeter:

. . \(\displaystyle \:4\:=\:4h\,+\,4\pi r\;\) . . . yes

. . \(\displaystyle \frac{4}{4\pi r}\:=\:4h\) . . . no

Of course, you meant: \(\displaystyle \:4h\,+\,4\pi r\:=\:4\;\;\Rightarrow\;\;4h\:=\:4\,-\,4\pi r\;\;\Rightarrow\;\;h\:=\:1\,-\,\pi r\)

Then: \(\displaystyle \:V \;= \; \pi r^2h \;=\;\pi r^2(1\,-\,\pi r)\) . . . etc.