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Math Help - Please help with optimization problem

  1. #1
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    Please help with optimization problem

    If 2100 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box.
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  2. #2
    Senior Member topher0805's Avatar
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    First find an equation to represent the surface area of the box. The surface area will be the surface area of the base plus the surface area of the sides:

    S_t=S_b+S_s

    The surface area of the base will be the width multiplied by the length, which in this case is just the width squared:

    S_b=w^2

    The surface area of the sides will be 4 multiplied by the surface area of one of the sides. The surface area of one of the sides is the width multiplied by the height:

    <br />
S_s=4wh

    <br />
S_t=w^2+4wh<br />

    We also know that the total surface area will be equal to 2100:

    2100=w^2+4wh

    h=\frac {2100-w^2}{4w}

    Volume is represented by the equation:

    <br />
V=l\cdot w\cdot h

    In this case, we know that l=w and we have already solved for h in terms of w, so we have that:

    V(w)=w^2\cdot \frac {2100-w^2}{4w}

    Now that you have the volume as a function of one variable, all you have to do is find the derivative and find the critical points. Once you have found the critical points, plug them back into the original equation to find the highest possible volume.
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  3. #3
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    When taking the derivative of this function...

    do you use the quotient rule inside the product rule and then factor out what you can?
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  4. #4
    Senior Member topher0805's Avatar
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    Quote Originally Posted by stepho_145 View Post
    do you use the quotient rule inside the product rule and then factor out what you can?
    Try simplifying the equation first:

    <br />
V(w)=w^2\cdot \frac {2100-w^2}{4w}<br />

    <br />
V(w)=\frac {2100w-w^3}{4}<br />

    <br />
V(w)=525w-\frac {1}{4}w^3<br />
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  5. #5
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    Thanks a lot...

    that was a big help.
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  6. #6
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    Still not getting the right answer

    Do you plug the critical values back into the original function? I'm getting 0, + and - 45.82575695 for my critical points.
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  7. #7
    Senior Member topher0805's Avatar
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    The critical values are the x values where the original function is at its max/mins. Plug them back into the original function, and the highest output will be the max.
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