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Math Help - largest possible volume of box

  1. #1
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    largest possible volume of box

    If 1900 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.
    so i tried doing this
    v=x^2y
    x^2+4xy=1900
    y=(1900-x^2)/4x
    =475x-x/4

    V=x^2(475x-x/4)
    taking the derivative and setting it to 0 i get
    12.58 but thats not right
    can anyone see what i did wrong
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  2. #2
    Senior Member eumyang's Avatar
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    If 1900 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.
    so i tried doing this
    v=x^2y
    x^2+4xy=1900
    y=(1900-x^2)/4x
    =475x-x/4
    The last step above is wrong.

    \begin{aligned}<br />
y &= \dfrac{1900 - x^2}{4x} \\<br />
&= \dfrac{1900}{4x} - \dfrac{x^2}{4x} \\<br />
&= \dfrac{475}{x} - \dfrac{x}{4}<br />
\end{aligned}

    So
    V = x^2 y = x^2 \left( \dfrac{475}{x} - \dfrac{x}{4} \right) = 475x -  \dfrac{x^3}{4}
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  3. #3
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    Unfortunately, you seem to have confused yourself. I will give you a great gift, My Friend. It is the gift of formality. You have gravely underestimated the value of being careful, of being explicit, and of being complete.

    1) You switched from 'v' to "V" in the middle of your demonstration. This may be only a typo, but it is confusing.
    2) You defined neither 'x' not 'y'. You should have done this carefully at the beginning. It takes very little time, but it will save you often.

    x = length of each side of the square base.
    y = height of the box.

    3) You did not explain your steps. Your second was very confusing. I first thought you were trying to find a derivative of v(x,y). It took a while to see what you were doing. WRITE IT DOWN and no one will have to be confused.

    Define the volume, v = y*x^2
    Define the surface area, one base and four equal sides: x^2 + 4xy = 1900

    4) You may wish to explore the use of Units. Units will save you, too. In this particularly simple problem, the units may be more confusing, but they will be necessary in a more complicated structure.

    Solve the Surface Area equation for 'y': y = (1900-x^2)/(4x)

    5) You forgot your Order of Operations. Without the parentheses in the denominator, you have it wrong.

    Simplify the expression for 'y': y = 475/x - x/4 -- Whoops! There's your problem. Your division turned into multiplication. I have to believe that the error in step 5 is the real problem.

    I'd go on a bit about not sharing your derivative expression and saying nothing of endpoints, but that's enough to swallow for one day.

    I hope you appreciate the gift. It is of great value.
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  4. #4
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    well i took the derivative of 475x-x^3/4 and got 475 - (3 x^2)/4 set that equal to 0 and got 25.17 however that is incorrect
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  5. #5
    Senior Member eumyang's Avatar
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    Of course it's incorrect! You found x ≈ 25.17, and x is the length/width of the square base. They're asking for the largest possible volume of the box! So how do we find it?
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  6. #6
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    I am saddened that you have not accepted your gift. Please reconsider your position.

    6) How did you manage 25.17?
    7) What is 25.17?

    The question asks for VOLUME. Why do you think you are done after finding in the length of the side? Oh, wait! I know!! Because you did not take the time to write careful and thorough definitions.

    In case I'm feeling a little lazy, I might try to be extra explicit:

    x = length of each side of the square base. When I find 'x', I am not yet done.
    y = height of the box. When I find 'y', I am not yet done.

    Define the volume, v = y*x^2 - The question asks for volume. When I find this, I'm done.

    In this case, however, you are not done. Are you SURE it's a maximum? REALLY SURE? Derivatives don't know the difference between minima and maxima. Derivatives don't work at endpoints, either. There is SO MUCH to think about. There just isn't time to be lazy or sloppy.
    Last edited by TKHunny; August 9th 2010 at 05:31 PM.
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  7. #7
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    ok i got the answer
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  8. #8
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    {sigh} But is it right?
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  9. #9
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    V = x^2 y
    <br />
A=x^2+4xy <br />
= \dfrac{V}{y}+\dfrac{4V}{x}<br />
= \dfrac{V}{y}+\dfrac{2V}{x}+\dfrac{2V}{x}<br />
\geq 3\sqrt[3]{\dfrac{4V^3}{x^2 y}}<br />
= 3\sqrt[3]{4V^2}<br />

    using AM-GM. Equality when x=2y
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