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Math Help - Maximize and Minimize Area

  1. #1
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    Maximize and Minimize Area

    Please help!!! I should know how to do this but it's just not happening for me.

    If you have 12 meters of wire that has to be cut into 2 pieces. 1 piece is to be bent into the shape of a circle and the other into the shape of a square. How should the wire be cut so as to: a) maximize the total area enclosed by the shapes. b) Minimize the total area enclosed by the shapes.

    Thanks for your help.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by Gigabyte View Post
    If you have 12 meters of the wire that has to be cut into 2 pieces. 1 piece is to be bent into the shape of a circle and the other into the shape of a square.
    Let's call L_{circ} the length of wire bent into the shape of a circle. L_{circ} is the perimeter of the circle and it is linked to its radius R by L_{circ}=2\pi R \implies R=\frac{L_{circ}}{2\pi}. Thus the area of the circle is A_{circ}=\pi R^2=\pi \frac{L_{circ}^2}{4\pi ^2}=\frac{L_{circ}^2}{4\pi}

    Let's do the same thing for the square : let's call L_{square} the length of wire bent into the shape of a square. L_{square} is the perimeter of the square thus the length of one of its side is \frac{L_{square}}{4} and its area is \left(\frac{L_{square}}{4}\right)^2= \frac{L_{square}^2}{4^2}

    As the length of wire is 12\mathrm{m}, L_{square}+L_{circle}=12 \implies L_{circle}=12-L_{square}. The area of the two shapes can be given in function of L_{square} :

    A=\frac{L_{square}^2}{4^2}+\frac{L_{circ}^2}{4\pi}  =\frac{L_{square}^2}{4^2}+\frac{(12-L_{square})^2}{4\pi}

    The problem boils down to finding the maximum and the minimum of the function A(x)=\frac{x^2}{4^2}+\frac{(12-x)^2}{4\pi} for x\in[0,\,12]. (because 0 \leq L_{square} \leq \L_{wire}=12 )

    Can you take it from here ?
    Last edited by flyingsquirrel; May 21st 2008 at 03:16 AM. Reason: area of the square
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  3. #3
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    thanks mate.

    But wouldn't the area of the square be
    (Lsquare/4)^2 not (Lsquare/4)^3?

    And then the differentiation of A i couldn't do. Well i've tried and i get x=6.53.
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Gigabyte View Post
    thanks mate.

    But wouldn't the area of the square be
    (Lsquare/4)^2 not (Lsquare/4)^3?
    Yes

    And then the differentiation of A i couldn't do. Well i've tried and i get x=6.53.
    \frac{\mathrm{d}A}{\mathrm{d}x}=2\frac{x}{4^2}-\frac{2(12-x)}{4\pi}=x\left(\frac{1}{8}+\frac{1}{2\pi}\right)-\frac{6}{\pi}=x\cdot\frac{\pi+4}{8\pi}-\frac{6}{\pi}

    This derivative equals 0 if x=\frac{48}{\pi+4}, it is negative for 0\leq x \leq \frac{48}{\pi+4} and positive if \frac{48}{\pi+4}\leq x \leq 12 hence A is increasing on \left[\frac{48}{\pi+4},\,12 \right] and decreasing on \left[0,\,\frac{48}{\pi+4} \right]. It can be deduced from this that the area is minimized at x_m=\frac{48}{\pi+4} and maximized at x_M=0 or x_M=12. To choose between this two values, one can compare A(0) and A(12) : x_M=0 if A(0)>A(12) and x_M=12 otherwise.
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