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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?

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

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

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