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I am really stupid and I can't get this on my own. If $1200 cm^2$ is used to make a square box with open top what is the largest volume the box could be?

I set it up like so $$v = l w h$$

$$v = w^2 h$$ $$1200 = w^2 + 4wh$$

Where do I go from here? Everything I try in wrong.

I tried to make it like so

$$h = 300/w 0 w/4$$ to eilimate h but it didn't help give an answer that is correct.

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I presume the constraint is area = 1,200 cm squared? –  copper.hat Jan 10 '13 at 21:25
    
Yes that is correct. –  user56699 Jan 11 '13 at 14:21

2 Answers 2

up vote 4 down vote accepted

If the surface area is fixed at $1200$ cm$^2$, then $$ 1200 = w^2 + 4wh \quad \Rightarrow\quad h = \frac{1}{4w}(1200 - w^2). $$ Then you can find the maximum volume from: $$ V = w^2h = w^2\frac{1}{4w}(1200 - w^2) = \frac{1}{4}w(1200 - w^2) $$


If the volume is fixed at $1200$ cm$^3$, then $$ 1200 = w^2h\quad \Rightarrow\quad h = \frac{1200}{w^2}. $$ Then you can find the minimum area from $$ A = w^2 + 4wh = w^2 + 4w\frac{1200}{w^2} = \dots $$

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There is a nice trick you can use here, to avoid calculus altogether: If you make two such boxes, you can turn one of them upside down and rest it on the other one, to get a closed box. This closed box will have the maximum volume for a surface area of 2400 $\text{cm}^3$, which is a cube of side 20. (You can prove this without calculus, just using the fact that for fixed $a$, $x(a-x)$ is maximised when $x=a/2$.)

So each open box has a volume of $20 \times 20 \times 10 = 4000 \text{cm}^3$.

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