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Math Help - Finding the cost for the cheapest container?

  1. #1
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    Finding the cost for the cheapest container?

    A rectangular container with no top has a volume of 20 m^3. The length of its base is twice the width.
    The cost for the material of the base is $5 per m^2.
    The cost for the material of the sides are $12 per m^2.
    What is the cost for the cheapest container?

    My attempted work:

    Let x be the width. Let y = length.
    V = 20 = xyz.

    C = 5xy + 12(2xz) + 12(2yz)
    C = 5xy + 24xz + 24yz

    V = 20 = xyz
    z = 20/(xy)

    C = 5xy + 240/x + 240/y

    Cx = 5y - 240/(x^2)
    Cy = 5x - 240/(y^2)

    Not sure how to get the answer. Can somebody show me how?
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  2. #2
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    Don't forget that the length is twice the width. That is, y=2x.

    So:

    C = 5xy + \frac{240}{x} + \frac{240}{y}

    \implies C = 5x(2x) + \frac{240}{x} + \frac{240}{2x}

    \implies C = 10x^2 + \frac{240}{x} + \frac{120}{x}

    Take the derivative from here.
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  3. #3
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    Quote Originally Posted by drumist View Post
    Don't forget that the length is twice the width. That is, y=2x.

    So:

    C = 5xy + \frac{240}{x} + \frac{240}{y}
    How did you get \frac{240}{x}

    and \frac{240}{y}
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  4. #4
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    Quote Originally Posted by ione View Post
    How did you get \frac{240}{x}

    and \frac{240}{y}
    I copied it from what AlphaRock showed, but it looks like he made a mistake.

    We have z=\frac{20}{xy} and y=2x, so:


    C = 5xy + 24xz + 24yz

    \implies C = 5xy + 24x\frac{20}{xy} + 24y\frac{20}{xy}

    \implies C = 5xy + \frac{480}{y} + \frac{480}{x}

    \implies C = 5x(2x) + \frac{480}{2x} + \frac{480}{x}

    \implies C = 10x^2 + \frac{240}{x} + \frac{480}{x}

    \implies C = 10x^2 + \frac{720}{x}
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