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Math Help - dimensions of box optimization

  1. #1
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    dimensions of box optimization

    how to find the dimensions of the rectangular box having the largest volume and surface area 150 square units?
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  2. #2
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    completely closed box or an open top box?
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  3. #3
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    completely closed...thankyou
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  4. #4
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    I was afraid you'd say that.

    The max volume for a closed box with a fixed surface area will be a cube.

    Here is the work ...

    let x , y , and z be the length, width, and height.

    V = xyz

    A = 150 = 2xy + 2xz + 2yz

    xy + xz + yz = 75

    z = \frac{75-xy}{x+y}

    V = \frac{75xy - x^2y^2}{x+y}

    \frac{\partial V}{\partial x} = \frac{(x+y)(75y - 2xy^2)-(75xy - x^2y^2)(1)}{(x+y)^2}

    \frac{\partial V}{\partial x} = \frac{75xy - 2x^2y^2 -2xy^3 + 75y^2 - 75xy + x^2y^2}{(x+y)^2}

    \frac{\partial V}{\partial x} = \frac{75y^2-x^2y^2-2xy^3}{(x+y)^2}

    since y \neq 0

    \frac{\partial V}{\partial x} = \frac{75-2xy-x^2}{(x+y)^2}


    \frac{\partial V}{\partial y} = \frac{(x+y)(75x - 2x^2y)-(75xy - x^2y^2)(1)}{(x+y)^2}

    \frac{\partial V}{\partial y} = \frac{75x^2 - 2x^3y + 75xy - 2x^2y^2 - 75xy + x^2y^2}{(x+y)^2}

    \frac{\partial V}{\partial y} = \frac{75x^2 - 2x^3y - x^2y^2}{(x+y)^2}

    since x \neq 0 ...

    \frac{\partial V}{\partial y} = \frac{75 - 2xy - y^2}{(x+y)^2}

    set both partials = 0 ...

    75-2xy-x^2 = 0

    75 - 2xy - y^2 = 0

    combine both equations ...

    x^2 - y^2 = 0

    (x-y)(x+y) = 0

    since x \neq -y ... x = y

    z = \frac{75-xy}{x+y} = \frac{75-x^2}{2x}

    V = x^2z = \frac{75x - x^3}{2}

    \frac{dV}{dx} = \frac{75 - 3x^2}{2} = 0

    x = 5

    y = 5

    z = 5

    the "cube"
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  5. #5
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    thankyou very much, i really appreciate it!!!
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