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Math Help - Optimization Problem

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    Optimization Problem

    If 1200 sq. cm of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

    I understand the basics of how these problems work, but applying that is difficult. Do I start with V= xy or something similar?
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by bobsanchez View Post
    If 1200 sq. cm of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

    I understand the basics of how these problems work, but applying that is difficult. Do I start with V= xy or something similar?
    The volume of the box is the area of the base times the height: V=x^2y.

    Note that since the box has an open top, the amount of material is the area of the 5 remaining sides: 1200=4xy+x^2

    Can you solve this problem now?
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    Um, I think so. Setting it up is definitely the hardest part for me. So from there I try to put it in terms of y and then differentiate, correct? Should that be y' = -(4x^2 - 4800)/(4x)^2 ?
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by bobsanchez View Post
    Um, I think so. Setting it up is definitely the hardest part for me. So from there I try to put it in terms of y and then differentiate, correct? Should that be y' = -(4x^2 - 4800)/(4x)^2 ?
    You need to plug y back into the volume equation, and then differentiate!!

    From the constraint, we see that y=\dfrac{1200-x^2}{4x}, which in return gives us the volume equation V=x^2\cdot\dfrac{1200-x^2}{4x}=300x-\tfrac{1}{4}x^3.

    Thus, V^{\prime}=\ldots. Set V^{\prime}=0 and solve for x.

    Can you proceed?
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    Ahhh that's what I've been doing wrong, okay. That makes much more sense. Let's see...I get V' = 300 - 3x^2/4, which when set equal to zero gives me x = 20...is this correct?
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by bobsanchez View Post
    Ahhh that's what I've been doing wrong, okay. That makes much more sense. Let's see...I get V' = 300 - 3x^2/4, which when set equal to zero gives me x = 20...is this correct?
    Correct!

    Now what is y?
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  7. #7
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    y still equals 300x - x^3/4.

    So... V = 4000?
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