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Thread: Minimizing Diagonals of rectangles

  1. #1
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    Minimizing Diagonals of rectangles

    Of all rectangles with given area, $\displaystyle A$, which has the shortest diagonals?

    I'm not sure what this question is asking, but I tried. I'll show my work, Please tell me if I did it correctly or help me if I didn't.

    x = length
    y = width
    A = area

    $\displaystyle A = xy$

    The diagonal would be $\displaystyle d = \sqrt{x^2 + y^2}$.

    Because I want to minimize the diagonals, I solved for y in the area function and substituted it into the formula for the diagonal.

    $\displaystyle d = \sqrt{x^2 + (\frac{A}{x})^2} $

    I'll take the derivative of the diagonal and set it equal to zero to minimize it (here i treated A as a constant... I'm not sure if i can do this.. this may be my mistake.. I'm not sure).

    $\displaystyle d' = \frac{1}{2} (x^2 + (\frac{A}{x})^2))^{-\frac{1}{2}}(2x - \frac{2A^2}{x^3}) = 0$
    $\displaystyle x - \frac{A^2}{x^3} = 0$
    $\displaystyle x^4 - A^2 = 0$
    $\displaystyle x^4 = A^2$
    $\displaystyle x = \sqrt{A}$

    Plug it back in to the area to find the other dimension.
    $\displaystyle A = \sqrt{A} * y$
    $\displaystyle y = \frac{A}{\sqrt{A}}$
    $\displaystyle y = \sqrt{A}$

    The dimensions should be $\displaystyle \sqrt{A} \quad x\quad \sqrt{A}$.

    Is this what they wanted? And is this correct? I have no clue =\
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  2. #2
    MHF Contributor Quick's Avatar
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    The answer is correct but I haven't looked through your work, notice that when a side is the square root of A then the rectangle is a square.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jacobpm64 View Post
    Of all rectangles with given area, $\displaystyle A$, which has the shortest diagonals?

    I'm not sure what this question is asking, but I tried. I'll show my work, Please tell me if I did it correctly or help me if I didn't.

    x = length
    y = width
    A = area

    $\displaystyle A = xy$

    The diagonal would be $\displaystyle d = \sqrt{x^2 + y^2}$.

    Because I want to minimize the diagonals, I solved for y in the area function and substituted it into the formula for the diagonal.

    $\displaystyle d = \sqrt{x^2 + (\frac{A}{x})^2} $

    I'll take the derivative of the diagonal and set it equal to zero to minimize it (here i treated A as a constant... I'm not sure if i can do this.. this may be my mistake.. I'm not sure).

    $\displaystyle d' = \frac{1}{2} (x^2 + (\frac{A}{x})^2))^{-\frac{1}{2}}(2x - \frac{2A^2}{x^3}) = 0$
    $\displaystyle x - \frac{A^2}{x^3} = 0$
    $\displaystyle x^4 - A^2 = 0$
    $\displaystyle x^4 = A^2$
    $\displaystyle x = \sqrt{A}$

    Plug it back in to the area to find the other dimension.
    $\displaystyle A = \sqrt{A} * y$
    $\displaystyle y = \frac{A}{\sqrt{A}}$
    $\displaystyle y = \sqrt{A}$

    The dimensions should be $\displaystyle \sqrt{A} \quad x\quad \sqrt{A}$.

    Is this what they wanted? And is this correct? I have no clue =\
    Yep, you got it right!

    -Dan
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