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Math Help - Perimeter word problem

  1. #1
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    Perimeter word problem

    Posted: Sun Nov 19, 2006 11:56 pm Post subject: Perimeter word problem

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    Hi,

    Im having some problems with a word problem given by my instructor.

    The quesiton is -

    You are given 50m of rope, and cut it into two sections (x and y). With one part of the rope, you make a circle. With the other half, you form a square.
    a) Create an equation to illustrate this idea and
    b) maximize the area of both objects

    Im having problems creating an equation for this problem. I know that

    x+y=50
    Peremeter of a circle - 2(Pi)r
    peremeter of a square - 4L

    Can anyone help me formulate the equation so I can maximize the peremeter?

    Thanks!

    Matt
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  2. #2
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    Quote Originally Posted by matt944 View Post
    Posted: Sun Nov 19, 2006 11:56 pm Post subject: Perimeter word problem

    --------------------------------------------------------------------------------

    Hi,

    Im having some problems with a word problem given by my instructor.

    The quesiton is -

    You are given 50m of rope, and cut it into two sections (x and y). With one part of the rope, you make a circle. With the other half, you form a square.
    a) Create an equation to illustrate this idea and
    b) maximize the area of both objects

    Im having problems creating an equation for this problem. I know that

    x+y=50
    Peremeter of a circle - 2(Pi)r
    peremeter of a square - 4L

    Can anyone help me formulate the equation so I can maximize the peremeter?

    Thanks!

    Matt
    the perimeter of the circle is x
    The perimeter of the square is y
    so P_x+P_y=50

    Now the trick is finding a relationship between perimeter and area, let's start with the circle.

    We have: P_x=2\pi r

    Divide both sides by 2: \frac{P_x}{2}=\pi r

    multiply both sides by r: \frac{P_x}{2}\cdot r=\pi r^2

    Substitute to get: \frac{P_x}{2}\cdot\frac{P_x}{2\pi}=A_x

    Thus: \frac{P_x^2}{4\pi}=A_x

    Now let's find the area of the square, we know that: P_y=4s where s means each side.

    Divide both sides by 4: \frac{P_y}{4}=s

    Square both sides: \frac{P_y^2}{16}=s^2

    Substitute: \frac{P_y^2}{16}=A_y

    Now go back to the equation: x+y=50

    Substitute: P_x+P_y=50

    Thus: P_y=50-P_x

    You are trying to maximize: A_x+A_y

    Substitute: \frac{P_x^2}{4\pi}+\frac{P_y^2}{16}

    Thus: \frac{4P_x^2}{16\pi}+\frac{P_y^2\pi}{16\pi}

    Add: \frac{4P_x^2+P_y^2\pi}{16\pi}

    Substitute: \frac{4P_x^2+(50-P_x)^2\pi}{16\pi}

    Thus: \frac{4P_x^2+\left(2500-100P_x+P_x^2\right)\pi}{16\pi}

    Then: \frac{4P_x^2+2500\pi-100P_x\pi+P_x^2\pi}{16\pi}

    Group like terms: \frac{4P_x^2+P_x^2\pi-100P_x\pi+2500\pi}{16\pi}

    Add: \frac{(4+\pi)P_x^2-100P_x\pi+2500\pi}{16\pi}

    Now maximize.
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