A piece of wire 13 m long is cut into two pieces.?

Question

A piece of wire 13 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. (Give your answers correct to two decimal places.)

(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? 5.65 m

I found B(and it is correct), I just cant find A for some odd reason, if you could show the steps and the answer that would be highly appreciated.

Answer 1

Let s = side of the square,
t = side of the triangle.
Then
4s + 3t = 13
A = s^2 + (1/4)t^2√3
t = (13 - 4s)/3
A = s^2 + (1/4)((13 - 4s)/3)^2√3
A = s^2 + ((169 - 104s + 16s^2)/36)√3
A = s^2 + 0.7698s^2 - 5.0037s + 8.1310
A = 1.7698s^2 - 5.0037s + 8.1310
dA/ds = 3.5396s - 5.0037 = 0 for minimum
s = 1.41 m
4s = 5.65 m
For maximum total area the square would use all the wire, 13m

Answer 2

The Extreme Value Theorem says that the extreme values on a closed interval occur either at a critical point or at an endpoint of the interval. So you need to consider the cases where all of the wire goes to making a square, or all to making an equilateral triangle, as well as the value you obtained from the critical point.