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|dw:1438823690584:dw|

is the length 2 or width 2?

no numbers there, just that

2w = length of box?

yes

is the box a rectangle as stated or is it a square

My attempt: Volume of the box is V = x*W*L = x*W (2W) = 2x W^2

They said the length is 2x bigger, so w = 1 in that case

I gotta run, continue with someone else

How can you find the volume if you know nothing about the height of the box?

Thanks

We consider Volume of the box is the function with respect to x and W

\(V = 2xW^2\\V'= 2W^2 + 2xW*W'\)

i think we're presuming maximum volume for a fixed perimeter, \(P=2(2x+w+2x+2w)\)

so $$P=8x+6w\implies x=\frac18(P-6w)$$

No, not that , the perimeter of what? if it is of the box, then you don't have x there, right?

oh, sorry, you are correct with perimeter. :)

which gives $$V^*=\frac14(P-6\cdot P/9)(P/9)^2=\frac14(P/3)(P^2/81)=\frac{P^3}{4\cdot 243}$$

I got what you meant. How about Surface area? how to work on it?

i get \[v(x) = 2 w^2 x-6 w x^2+4 x^3\]does that help?

and also \(x=\frac18(P-6w)=\frac18(P/3)=\frac{P}{24}\)

yeah take the derivative wrt x get\[2 (w^2-6 w x+6 x^2)\]

set equal to zero, solve for \(x\) get something like
\[(\frac{1}{2}+\frac{\sqrt3}{6})w\]

or maybe that is the wrong one

@misty1212 because V = 2xW^2 , not as your equation.

\[(\frac{1}{2}-\frac{\sqrt3}{6})w\] is probably better

hmm i get
\[V=(w-2x)(2w-2x)x\] as a start

multiply out get
\[V(x)=2 w^2 x-6 w x^2+4 x^3\]

@misty1212 not that, the w is the width of the box, 2w is the length of the box and x is its height

not the way you drew it for sure