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Optimization Problem involving a wire

  1. Apr 20, 2013 #1
    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. [itex]A_{s}(x) = x^{2}, A_{t}(x)=\frac{\sqrt{3}}{4}x^{2}[/itex]



    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.[itex]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}}[/itex] How do I simplify this?
     
  2. jcsd
  3. Apr 20, 2013 #2

    SteamKing

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    Arithmetic is a good start. Can you find a common denominator for the denominator?
     
  4. Apr 20, 2013 #3
    If I simplify it, here's how it looks, [itex]\frac{20\sqrt{3}}{9+4\sqrt{3}}[/itex] and if I multiply by 2, I get[itex]\frac{40\sqrt{3}}{9+4\sqrt{3}}[/itex]. Does this look right?
     
  5. Apr 21, 2013 #4

    SteamKing

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    That looks OK.
     
  6. Apr 21, 2013 #5

    haruspex

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