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Thread: Optimization problem, perimeter of a rectangle

  1. #1
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    Optimization problem, perimeter of a rectangle

    Dear forum members,

    A problem asks me to show that a rectangle of constant area A has a minimal perimeter when the rectangle is a square.

    I make the equation for the area and differentiate it with respect to x.

    $\displaystyle P’(x)= 2-\frac{2A}{x^2}$

    Solving for the zero points

    $\displaystyle 2x^2-2A=0 $

    At this stage, am I allowed to substitute A=xh into the equation and solve it as
    $\displaystyle 2x^2-2xh=0$

    $\displaystyle 2x(x-h)=0$
    x=0 or x=h

    or should I solve it as

    $\displaystyle x=-\sqrt{A}$ or $\displaystyle x=\sqrt{A}$

    because evidently the answer is different .
    Please help me, I am really confused.
    Thank you in advance.
    Last edited by Coach; Apr 7th 2009 at 11:49 AM.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by Coach View Post
    ...show that a rectangle of constant area A has a minimal perimeter when the rectangle is a square.
    I'm not sure what your initial equations were or what you were doing with them...?

    Try starting with the "area" and "perimeter" formulas for a rectangle with height "h" and width "w".

    . . . . .$\displaystyle A\, =\, hw$

    . . . . .$\displaystyle P\, =\, 2h\, +\, 2w$

    Solving the "area" formula for one of the dimensions (keeping in mind that the area, in this case, is a constant value), you get:

    . . . . .$\displaystyle h\, =\, \frac{A}{w}$

    Subsitituting, you get:

    . . . . .$\displaystyle P(w)\, =\, \frac{2A}{w}\, +\, 2w$

    Differentiating, and keeping in mind that dA/dw = 0, you get:

    . . . . .$\displaystyle P'(w)\, =\, \frac{2\frac{dA}{dw}w\, -\, 2A}{w^2}\, +\, 2\, =\, -\frac{2A}{w^2}\, +\, 2$

    Setting this equal to zero, you get:

    . . . . .$\displaystyle -\frac{2A}{w^2}\, +\, 2\, =\, 0$

    . . . . .$\displaystyle -2A\, +\, 2w^2\, =\, 0$

    . . . . .$\displaystyle w^2\, =\, A$

    . . . . .$\displaystyle w\, =\, \sqrt{A}$

    (Since the width cannot be negative, you can ignore the negative square root.)

    Since:

    . . . . .$\displaystyle A\, =\, hw$

    ...and since the value you want is at w = sqrt[A], then:

    . . . . .$\displaystyle A\, =\, h\sqrt{A}$

    ...then:

    . . . . .$\displaystyle h\, =\, \frac{A}{\sqrt{A}}\, =\, \sqrt{A}$

    What sort of rectangle has the same width as height?
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