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?
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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?
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.
Steve, I had my definitions reversed, but
http://www.jiskha.com/display.cgi?id=1384823309
Only superficially checked the arithmetic on both
Well, yeah, that and the fact that your wire was 40m long, not 23.
Well, I simply offered another approach.
I would have expected you to change the 40 to 23.