An open box with the length is twice its width. Find the maximum value that the volume of the box can have. Please, help.

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An open box with the length is twice its width. Find the maximum value that the volume of the box can have. Please, help.

Mathematics
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|dw:1438823690584:dw|
is the length 2 or width 2?
no numbers there, just that

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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
Surface are of the open box is \(S = (L+2x)(W+2x) - 4x^2= (2W+2x)(W+2x)-4x^2\) \(S = 2W^2+4xW+2xW+4x^2-4x^2= 2W^2+6xW\)
I gotta run, continue with someone else
How can you find the volume if you know nothing about the height of the box?
Thanks
@Mer I thought the same, anyways np loser.
@Mertsj maximum of volume at the point of derivative of the Volume =0
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?
|dw:1438825047689:dw| that is perimeter. One more thing, the prof asked us to work with Surface area. I am sorry for this information
fixed perimeter of the original rectangle that we cut the box out of. so now we have $$V(x,w)=w\cdot2w\cdot x=2xw^2$$now using \(x=\frac18(P-6w)\) we have: $$V=\frac14(P-6w)w^2=\frac14(Pw^2-6w^3)$$now take: $$\frac{dV}{dw}=\frac14(2Pw-18w^2)=\frac12(Pw-9w^2)$$so $$\frac{dV}{dw}=0\\Pw-9w^2=0\\w(P-9w)=0\implies w=P/9$$
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 \(x=\frac18(P-6w)=\frac18(P/3)=\frac{P}{24}\)
@misty we have to find x such that the Volume is maximum @oldrin.bataku Again, I have to work with Surface area, not perimeter. Please
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
oh i see then the way the question is asked, there is no maximum make w bigger and bigger the volume increases without bound
@oldrin.bataku I saw you wrote something, where is it?? why don't you post it?
so for a fixed surface area? $$A=2w^2+2wx+2(2w)x\\\quad =2w^2+6wx$$so we have \(x=\frac{A-2w^2}{6w}=\frac{A}{6w}-\frac{w}3\) so $$V=2xw^2=2\left(\frac{A}{6w}-\frac{w}3\right)w^2=\frac{Aw}{3}-\frac{2w^3}{3}\\\frac{dV}{dw}=\frac{A}3-2w^2$$so $$\frac{A}3-2w^2=0\\2w^2=A/3\\w^2=A/6\\w=\sqrt{A/6}$$
Where is your x? is it not that we have to find the ratio of x and W such that V is max? I meant like x = 1/5 W, then V is max.
plug it in: $$x=\frac{A-2\cdot A/6}{6\cdot\sqrt{A/6}}=\frac{A-A/3}{\sqrt{6A}}=\frac{2A/3}{\sqrt{6A}}=\frac{\sqrt{2A}}{3\sqrt3}$$
so the ratio of w, x is just $$x/w=\frac{\sqrt{2A}\cdot\sqrt6}{3\sqrt3\cdot\sqrt{A}}=\frac{\sqrt2\cdot\sqrt2}{3}=\frac23$$
I got exactly the same, but it is impossible when \(0
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