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Math Help - Optimization Problem3

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    Optimization Problem3

    Rectangle storage container with an open top is to have a volume 10 m^3.
    The length of its base is twice the width.
    Material for the base costs $10 /m^2.
    Material for the sides costs $6 /m^2.
    Find the cost of materials for the cheapest such container.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by camherokid View Post
    Rectangle storage container with an open top is to have a volume 10 m^3.
    The length of its base is twice the width.
    Material for the base costs $10 /m^2.
    Material for the sides costs $6 /m^2.
    Find the cost of materials for the cheapest such container.
    let the width of the box be x, then the length is 2x
    let the height of the box be y

    we want a box of the smallest surface area to minimize the cost.

    we have V = lwh
    => V = x(2x)(y) = 2x^2 * y
    we are told V = 10
    => 2x^2 * y = 10
    => y = 10/2x^2 = 5/x^2

    now the surface area is given by
    S = area of base + area of the sides
    => S = 2x^2 + 6xy
    => S = 2x^2 + 6x(5/x^2)
    => S = 2x^2 + 30x^-1

    => S' = 4x -30x^-2
    for min surface area, set S' = 0
    => 4x - 30x^-2 = 0
    => 4x^3 - 30 = 0 .............multiplied through by x^2
    => 4x^3 = 30
    => x^3 = 30/4 = 15/2
    => x = cuberoot(15/2) ~= 1.9574

    but y = 5/x^2 = 5/(1.9574)^2 ~= 1.305

    so for min surface area: x = 1.9574, y = 1.305

    so the area of the base = 2x^2 = 7.6628
    so the cost of the base is $76.63

    so the area of the sides is 6xy = 6(1.9574)(1.305) = 15.3264
    so the cost of the sides is $91.96

    so the cost of the material needed to make such a box is $91.96 + $76.63 = $168.59 ~= $169
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