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anonymous

  • 5 years ago

If 3072 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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  1. anonymous
    • 5 years ago
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    Let l be length, w the width and h, the height. Then, you have the following pieces of information: (1) Surface area of open box is \[A=lw+2wh+2lh=w^2+4wh\] since the base of the box is square, l=w. You're adding up the areas of the faces (except for the top since it's open). (2) The volume, which is ultimately what you want to find,\[V=lwh=w^2h\]Now, you want to find the dimensions that maximize the volume. The problem you have is that there are two dimensions here, w and h. To take a full, explicit derivative in one variable, you need to have everything in only that variable. But we're in luck, since you have another expression that relates how height should vary with w - the relationship concerning area. So, from area, you can solve for h in terms of w to get,\[h=\frac{A-w^2}{4w}\]Substituting this into the volume expression,\[V=w^2\left( \frac{A-w^2}{4w} \right)=\frac{Aw}{4}-\frac{w^3}{4}\]Then

  2. anonymous
    • 5 years ago
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    \[\frac{dV}{dw}=\frac{A}{4}-\frac{3w^2}{4}\]This will yield w extreme in the interval of concern for \[\frac{dV}{dw}=0\]that is, when\[\frac{A-3w^2}{4}=0 \rightarrow w=\sqrt[3]{\frac{A}{3}}=\sqrt[3]{\frac{3072}{3}}\approx 10cm\] The height can then be found from\[h=\frac{A-w^2}{4w}\approx \frac{3072-100}{400}=7.43cm\]

  3. anonymous
    • 5 years ago
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    Now, at this point, you really need to take the second derivative to prove that the point you found does indeed yield a maximum. Then \[\frac{d^2V}{dw^2}=-\frac{3w}{2}\]which is less than zero when w is positive. So the w we have here will yield a maximum. So, the maximum volume is given by \[V_{maximum}=w^2happrox 10^2 \times 7.43=743\]cm^3.

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