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Thread: Extreme Value Problems

  1. #1
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    Extreme Value Problems

    A box with an open top is to be constructed from a square piece of cardboard, 6 meters wide, by cutting out a square from each of the four corners, and bending up the sides. Find the largest volume that such a box can have.



    Let x represent the dimensions

    $\displaystyle V = (6 - 2x)^2x$
    $\displaystyle V = 36x - 24x^2 + 4x^{3}$

    $\displaystyle V' = 36 - 48x + 12x^2$
    $\displaystyle 0 = 36 - 48x + 12x^2$
    $\displaystyle 0 = 12(x - 1)(x - 3)$

    $\displaystyle x = 1 and x = 3$


    Restrictions: $\displaystyle 0 \leq x \leq 3$

    $\displaystyle V = Lwh$
    $\displaystyle V = (3)(1)(6)$
    $\displaystyle V = 18m^{3}$

    ---------

    Text Answer is $\displaystyle 16m^{3}$

    Please help me correct my work?
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  2. #2
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    Quote Originally Posted by Macleef View Post
    A box with an open top is to be constructed from a square piece of cardboard, 6 meters wide, by cutting out a square from each of the four corners, and bending up the sides. Find the largest volume that such a box can have.



    Let x represent the dimensions

    $\displaystyle V = (6 - 2x)^2x$
    $\displaystyle V = 36x - 24x^2 + 4x^{3}$

    $\displaystyle V' = 36 - 48x + 12x^2$
    $\displaystyle 0 = 36 - 48x + 12x^2$
    $\displaystyle 0 = 12(x - 1)(x - 3)$

    $\displaystyle x = 1 and x = 3$


    Restrictions: $\displaystyle 0 \leq x \leq 3$

    $\displaystyle V = Lwh$
    $\displaystyle V = (3)(1)(6)$
    $\displaystyle V = 18m^{3}$

    ---------

    Text Answer is $\displaystyle 16m^{3}$

    Please help me correct my work?
    You need to plug into the volume formula

    $\displaystyle V(x) = (6 - 2x)^2x$

    $\displaystyle V(1) = (6 - 2(1))^2(1)=(4)^2=16$
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