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I am suppose to find the volume if 1200 cm^2 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 think what I need to do is set up the formulas to be

$4(lw) + w^2 = 1200$ for area

$lwh = v$ for volume

I know that if the base is a square than the rectangle will have the same dimensions and the only different variable would be the height so I can solve for length like so

$l=\frac{1200-w^2}{4w}$

Now that I have that I can put it in my formula

$lwh = v$ for volume

which I can rewrite as

$l^2 * h = v$

I then take the derivative of this and I get some ridiculous answer that is wrong.

$\frac{1200w^2 - w^4}{4w}$

the derivative

$300 \frac{-3w^2}{4}$

Which gives me $+-20$ which is an incorrect answer.

2 Answers 2

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The formula $\rm V=lwh$ means "volume = length times width times height." The variable $\rm l$ is length, the variable $\rm w$ is width, and the variable $\rm h$ is height. Using these, the total area is actually

$\rm 2(l\times h)+2(w\times h)+w^2=1200.$

We know that $\rm l=w$ (because the base of the box is square), so this is $\rm 4wh+w^2=1200$. This allows us to solve for the height $\rm h$ in terms of width $\rm w$ as $\rm h=(1200-w^2)/(4w)$.

We have the formula for $\rm h$ in terms of $\rm w$, and know $\rm l=w$, so we have the volume function

$\rm V=l\,wh=w^2\frac{1200-w^2}{4w}=300w-\frac{1}{4}w^3.$

Now can you take the derivative of this, equate it with zero and solve for $\rm w$?

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    @Jordan: As per the [derivative test](http://en.wikipedia.org/wiki/First_derivative_test) (and a couple other checks), this means the volume is optimized when the width is twenty. Plug $\rm w=20$ into the formula for $V$ and you will have the optimized volume.2012-04-02
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When you write down a formula, you must write down what the letters stand for.

When you write down $4lw+w^2=1200$, you must add, "where $l$ is the length of the base of the box, and $w$ is the width of the base of the box".

If you do that, you might see right away where you've messed up. You've written down a formula for the area which doesn't include the height of the box---that can't possibly be right, can it?

In fact, the formula you have written down only makes sense if $l$ is the height of the box, right?

So go back and identify the variables explicitly and then write down formulas that make sense.

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    @robjohn (or anon, or whatever), I take your point. I propose to deal with one issue at a time.2012-04-02