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Math Help - Minimum

  1. #1
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    Minimum

    A 384 - square meter plot of land is to be enclosed by a fence and divide into two equal parts by another fence parallel to one pair of sides. What dimensions of the outer rectangle will minimize the amount of fence used?
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  2. #2
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    Quote Originally Posted by deltaX View Post
    A 384 - square meter plot of land is to be enclosed by a fence and divide into two equal parts by another fence parallel to one pair of sides. What dimensions of the outer rectangle will minimize the amount of fence used?
    Take the sides to which the inner fence is parallel to as having length x.
    Then the other two sides have length y.

    Fence length = 3x+2y

    Area =xy=384

    Substitute x or y into the linear equation

    xy=384\ \Rightarrow\ x=\frac{384}{y}

    \frac{3(384)}{y}+2y=k

    1152+2y^2=ky

    Differentiate and equate to zero to find minimum y

    4y=k

    1152=2y^2

    y=\sqrt{576}=24

    x=\frac{384}{24}=16
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  3. #3
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    Quote Originally Posted by Archie Meade View Post
    Take the sides to which the inner fence is parallel to as having length x.
    Then the other two sides have length y.

    Fence length = 3x+2y

    Area =xy=384

    Substitute x or y into the linear equation

    xy=384\ \Rightarrow\ x=\frac{384}{y}

    \frac{3(384)}{y}+2y=k

    1152+2y^2=ky

    Differentiate and equate to zero to find minimum y

    4y=k

    1152=2y^2

    y=\sqrt{576}=24

    x=\frac{384}{24}=16

    I was just somehow confused on one equation that is: 1152=2y^2

    where did that come from?
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  4. #4
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    Hi deltax,

    sorry, that was a few steps in one go!

    \frac{3(384)}{y}+2y=k

    \frac{1152}{y}+2y=k

    multiplying by y

    1152+2y^2=ky (1)

    Differentiate

    4y=k

    Now, use this instead of k in (1)

    1152+2y^2=4y(y)=4y^2

    Subtract 2y^2 from both sides

    1152+2y^2-2y^2=4y^2-2y^2

    1152=2y^2

    y^2=\frac{1152}{2}=576

    y=\sqrt{576}
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