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

- Loser66

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- Loser66

|dw:1438823690584:dw|

- anonymous

is the length 2 or width 2?

- Loser66

no numbers there, just that

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## More answers

- anonymous

2w = length of box?

- Loser66

yes

- anonymous

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

- Loser66

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

- anonymous

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

- Loser66

Surface are of the open box is \(S = (L+2x)(W+2x) - 4x^2= (2W+2x)(W+2x)-4x^2\)
\(S = 2W^2+4xW+2xW+4x^2-4x^2= 2W^2+6xW\)

- anonymous

I gotta run, continue with someone else

- Mertsj

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

- Loser66

Thanks

- anonymous

@Mer I thought the same, anyways np loser.

- Loser66

@Mertsj maximum of volume at the point of derivative of the Volume =0

- Loser66

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

- Loser66

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

- anonymous

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

- anonymous

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

- Loser66

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

- Loser66

|dw:1438825047689:dw| that is perimeter.
One more thing, the prof asked us to work with Surface area. I am sorry for this information

- anonymous

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 $$\frac{dV}{dw}=0\\Pw-9w^2=0\\w(P-9w)=0\implies w=P/9$$

- Loser66

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

- anonymous

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

- Loser66

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

- misty1212

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

- anonymous

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

- Loser66

@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

- misty1212

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

- misty1212

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

- misty1212

or maybe that is the wrong one

- Loser66

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

- misty1212

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

- misty1212

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

- misty1212

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

- Loser66

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

- misty1212

not the way you drew it for sure

- misty1212

oh i see
then the way the question is asked, there is no maximum
make w bigger and bigger
the volume increases without bound

- Loser66

@oldrin.bataku I saw you wrote something, where is it?? why don't you post it?

- anonymous

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 $$\frac{A}3-2w^2=0\\2w^2=A/3\\w^2=A/6\\w=\sqrt{A/6}$$

- Loser66

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.

- anonymous

plug it in: $$x=\frac{A-2\cdot A/6}{6\cdot\sqrt{A/6}}=\frac{A-A/3}{\sqrt{6A}}=\frac{2A/3}{\sqrt{6A}}=\frac{\sqrt{2A}}{3\sqrt3}$$

- anonymous

so the ratio of w, x is just $$x/w=\frac{\sqrt{2A}\cdot\sqrt6}{3\sqrt3\cdot\sqrt{A}}=\frac{\sqrt2\cdot\sqrt2}{3}=\frac23$$

- Loser66

I got exactly the same, but it is impossible when \(0

- Loser66

|dw:1438827425494:dw|

- Loser66

can't, not can

- anonymous

so use the extreme value theorem

- anonymous

if its not a local maximum in our domain, its an extreme point of our domain

- Loser66

Please, show me. I am struggle with it.

- anonymous

well, our domain has to be compact for the extreme value theorem, so we must look at \(x\in[0,w/2]\)

- Loser66

yes. That is the problem.

- anonymous

so $$x=0\implies A-2w^2=0\implies w=\sqrt{A/2}$$and $$x=w/2\implies 6w\cdot w/2=A-2w^2\implies w=\sqrt{A/5}$$

- anonymous

i think the problem is that these are physically unrealizable, though, and unfortunately the surface area constraint is just not strong enough

- anonymous

unless we've made some error

- Loser66

No, I went my way and work with derivative of surface area, I got exactly that x = 2/3 W
plug in, V = 2x W^2 = 4/3 W ^3
let's take x = W/2, V = 2 (W/2)W^2 = W^3 < 4/3 W^3, right?
testing more, we get V smaller and smaller , hence x =2/3 W seems right, but it is not right at all since it is out of domain. HA !!!

- Loser66

Let me show you my way
\(V = 2xW^2\\V'= 2W^2+ 4xW*W'\)
From \(A= 2W^2+6xW\\A'=0 = 4W*W' +6W+6xW'\)
That gives \(W'=\dfrac{-6W}{4W+6x}\)
replace to V' =0
\(2W^2+4xW\dfrac{-6W}{4W+6x}=0\)
\(2W^2 =\dfrac{24xW^2}{4W+6x}\)
\(12x = 4W +6x\\x = \dfrac{4}{6}W=\dfrac{2}{3}W\)

- anonymous

Actually, my way is definitely valid. the constraint to be less than $w/2$ is incorrect. It only must be that $x\ge0, w\ge0$. You are thinking of the constraint that $x$ must be less than half of the TOTAL side length $2x + w$, which is trivially true.

- Loser66

Algebraic way, we get stationary point is at x =(2/3)W, but it is out of the domain. I don't know how to do next.

- Loser66

ohhhhhhhhhhhhhh yes!!! I got you.

- Loser66

So, we are correct, since w is the width of the box, not the width of the paper. aaaaaaaaaaaaaahhhaaaaha... we got it right.... Thank you so so so so much.

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