# Optimization

• Jun 19th 2006, 11:28 AM
Jabés
Optimization
A piece of wire long L is cut in two pieces. One piece is bent into a semicircle and the other is bent into a equilateral triangle. How should the wire be cut so that the total area enclosed is (a) a maximun and (b) a minimun?

I'm stuck in this exercise. I dont know what to do :confused:
• Jun 19th 2006, 12:44 PM
ticbol
Quote:

Originally Posted by Jabés
A piece of wire long L is cut in two pieces. One piece is bent into a semicircle and the other is bent into a equilateral triangle. How should the wire be cut so that the total area enclosed is (a) a maximun and (b) a minimun?

I'm stuck in this exercise. I dont know what to do :confused:

Here is one way.

Cut the L long wire into x and (L-x) pieces, where x is in terms of L.

Bend the x into a semi-cirle. The (L-x) into an equilateral triangle.
(Let us say you can do that.)

Area enclosed by x:
Circumference = x = [2pi(r)]/2 +2r = r(pi +2)
r = x/(2+pi) = (0.1945)x
Area1 = [pi(r^2)]/2 = (1/2)*pi*[(0.1945)x]^2 = (0.0594)(x^2) ----***

Area enclosed by (L-x):
Each side of the equilateral triangle is (L-x)/3
Area = (1/2)(a)(b)sinC
Area2 = (1/2)[(L-x)/3][(L-x)/3]sin(60deg)
Area2 = (sqrt(3) /36)[(L-x)^2] = (0.0481)*(L-x)^2 -------------***

Total area enclosed, A, is:
A = (0.0594)(x^2) +(0.0481)[(L-x)^2]

For max or min A, set dA/dx to zero.
dA/dx = (0.1188)x +(0.0962)(L-x)(-1)
dA/dx = (0.1188)x +(0.0962)x -(0.0962)L

(I accidentally pressed the send button)

To continue...
dA/dx = (0.215)x -(0.0962)L
Set that to zero,
x = (0.0962)L /(0.215) = (0.4474)L

Therefore, for maximum enclosed areas, cut the L long wire into (0.4474)L and (0.5526)L pieces. -------------answer.

For minimum enclosed areas?
I guess cut the L long wire into the same pieces as above, but make semi-circle out of the (0.5526)L piece and equilateral triangle out of the (0.4474)L piece.
• Jun 19th 2006, 12:47 PM
Jabés