# Minimize total area of a square and triangle made of 13m long wire

I'm a little bit confused about this problem. I've gotten the first part, but I can't get the second!

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.

(a) How much wire should be used for the square in order
to maximize the total     area?

for this I got 13m

(b) How much wire should be used for the square in order to
minimize the total area?

Having trouble with this one. I keep getting:


$$(53sqrt(3))/(9+4sqrt(3))$$

but the online program that gave me the assignment is saying this is wrong. Any idea what I'm doing wrong?

• check your calculation again, should be 52 instead of 53. Apr 18 '14 at 1:11

The function you seek to minimize is

$$f(x) = \frac{\sqrt{3}}{4} \left (\frac{x}{3} \right )^2 + \left (\frac{13-x}{4} \right )^2$$

Then

$$f'(x) =\frac{\sqrt{3} x}{18} - \frac{13-x}{8} = \left (\frac{\sqrt{3}}{18}+\frac18 \right )x - \frac{13}{8}$$

Note that $f''(x) \gt 0$ so that the critical point at $f'(x)=0$ will be a minimum. The critical point is at

$$x=\frac{117}{9 + 4 \sqrt{3}} \approx 7.345 \, \text{m}$$

So that the amount used for the square will be $13-x$, or

$$13-x = \frac{52}{4+3 \sqrt{3}} \approx 5.655 \, \text{m}$$