Asked by TayB

A piece of wire 23 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.
(a) How much wire should be used for the square in order to maximize the total area?
(b) How much wire should be used for the square in order to minimize the total area?
Answered by Anonymous
see related questions below
Answered by TayB
Anonymous. That did not help at all. I already know part a is 23 m. For part b i had said 10 m, but got that marked wrong. Anybody know how to get what part b is?
Answered by Steve
If the triangle has side x, and the square has side y, then

3x+4y = 23
The area is
a = √3/4x^2 + y^2
= √3/4 x^2 + (23-3x)^2/16
= 1/16 (9+4√3)x^2 - 138x + 529)

so, for max/min area, we need da/dx=0, or
(18+8√3)x - 138 = 0
x = 69/(9+4√3) = 4.332
So, that means we use 13.0 for the triangle, and 10.0 for the square.

Not sure why you were marked off.
Answered by Reiny
Steve, I had my definitions reversed, but

http://www.jiskha.com/display.cgi?id=1384823309

Only superficially checked the arithmetic on both
Answered by @Reiny
Well, yeah, that and the fact that your wire was 40m long, not 23.
Answered by Reiny
Well, I simply offered another approach.
I would have expected you to change the 40 to 23.
Answered by TayB
i was marked off because it wanted exact answer and the exact answer for part b was (92sqrt(3))/(9+4sqrt(3)) it which is 10.00......
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