# Optimization Problem involving a wire

1. Apr 20, 2013

### frosty8688

1. A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is (a) a maximum? (b) A minimum?

2. $A_{s}(x) = x^{2}, A_{t}(x)=\frac{\sqrt{3}}{4}x^{2}$

3. The first equation is the length of the side of a square and the second is the length of a side of an equilateral triangle and the height of the triangle.$L_{s}=\frac{x}{4}, L_{s2}=\frac{10-x}{3}, h=\frac{\sqrt{3}}{6} (10-x), A(x)=\frac{x^{2}}{16}+\frac{1}{6}(10-x)*\frac{\sqrt{3}}{6}(10-x) = \frac{x^{2}}{16}+\frac{\sqrt{3}}{36}(10-x)^{2}; A'(x)=\frac{x}{8}-\frac{\sqrt{3}}{18}(10-x); x= \frac{\frac{5\sqrt{3}}{9}}{\frac{1}{8}+\frac{\sqrt{3}}{18}}=\frac{\frac{5\sqrt{3}}{9}}{\frac{1}{4}+\frac{\sqrt{3}}{9}}$ How do I simplify this?

2. Apr 20, 2013

### SteamKing

Staff Emeritus
Arithmetic is a good start. Can you find a common denominator for the denominator?

3. Apr 20, 2013

### frosty8688

If I simplify it, here's how it looks, $\frac{20\sqrt{3}}{9+4\sqrt{3}}$ and if I multiply by 2, I get$\frac{40\sqrt{3}}{9+4\sqrt{3}}$. Does this look right?

4. Apr 21, 2013

### SteamKing

Staff Emeritus
That looks OK.

5. Apr 21, 2013

### haruspex

You can simplify that a bit more. Do you know a trick for getting rid of terms like a+√b from denominators?