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Math Help - Differentiation Maximization Word Problem.

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    Differentiation Maximization Word Problem.

    A rectangular sheet of cardboard, 50 cm by 40 cm, is to have square corners cut of so it can be folden into a rectangular tray.

    Find the maximum volume possible for the tray.

    Here is what I got so far.

    Volume of Tray: L x W x H
    Volume of Tray with variable: L=(50 - 2x) x W(40 - 2x) x H=x
    Where x is equal to the cut out squares.
    (50 - 2x)(40 - 2x)x
    (2000 - 100x - 80x +4x^2)x
    2000x - 100x^2 - 80x^2 + 4x^3
    F'(x) = 2000 - 40x + 12x^2
    4(500 - 40x + 3x^2)

    Ahaha, I'm not sure what is going wrong...
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  2. #2
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    Quote Originally Posted by Liparulo View Post
    A rectangular sheet of cardboard, 50 cm by 40 cm, is to have square corners cut of so it can be folden into a rectangular tray.

    Find the maximum volume possible for the tray.

    Here is what I got so far.

    Volume of Tray: L x W x H
    Volume of Tray with variable: L=(50 - 2x) x W(40 - 2x) x H=x
    Where x is equal to the cut out squares.
    (50 - 2x)(40 - 2x)x
    (2000 - 100x - 80x +4x^2)x
    2000x - 100x^2 - 80x^2 + 4x^3
    F'(x) = 2000 - 40x + 12x^2
    4(500 - 40x + 3x^2)

    Ahaha, I'm not sure what is going wrong...
    Looks good so far.

    You have V = (50 - 2x)(40 - 2x)x

     = 2000x - 180x^2 + 4x^3.


    To find the maximum, you have to differentiate, set equal to 0 and solve for x.

    \frac{dV}{dx} = 2000 - 360x + 12x^2

    12x^2 - 360x + 2000 = 0

    Solve for x.
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