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Loser66
 one year ago
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
 one year ago
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is the length 2 or width 2?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0no numbers there, just that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is the box a rectangle as stated or is it a square

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0My attempt: Volume of the box is V = x*W*L = x*W (2W) = 2x W^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0They said the length is 2x bigger, so w = 1 in that case

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Surface are of the open box is \(S = (L+2x)(W+2x)  4x^2= (2W+2x)(W+2x)4x^2\) \(S = 2W^2+4xW+2xW+4x^24x^2= 2W^2+6xW\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I gotta run, continue with someone else

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.0How can you find the volume if you know nothing about the height of the box?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Mer I thought the same, anyways np loser.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@Mertsj maximum of volume at the point of derivative of the Volume =0

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0We consider Volume of the box is the function with respect to x and W

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0\(V = 2xW^2\\V'= 2W^2 + 2xW*W'\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think we're presuming maximum volume for a fixed perimeter, \(P=2(2x+w+2x+2w)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so $$P=8x+6w\implies x=\frac18(P6w)$$

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0No, not that , the perimeter of what? if it is of the box, then you don't have x there, right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438825047689:dw that is perimeter. One more thing, the prof asked us to work with Surface area. I am sorry for this information

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0fixed 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 \(x=\frac18(P6w)\) we have: $$V=\frac14(P6w)w^2=\frac14(Pw^26w^3)$$now take: $$\frac{dV}{dw}=\frac14(2Pw18w^2)=\frac12(Pw9w^2)$$so $$\frac{dV}{dw}=0\\Pw9w^2=0\\w(P9w)=0\implies w=P/9$$

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0oh, sorry, you are correct with perimeter. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which gives $$V^*=\frac14(P6\cdot P/9)(P/9)^2=\frac14(P/3)(P^2/81)=\frac{P^3}{4\cdot 243}$$

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I got what you meant. How about Surface area? how to work on it?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0i get \[v(x) = 2 w^2 x6 w x^2+4 x^3\]does that help?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and also \(x=\frac18(P6w)=\frac18(P/3)=\frac{P}{24}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.0yeah take the derivative wrt x get\[2 (w^26 w x+6 x^2)\]

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0set equal to zero, solve for \(x\) get something like \[(\frac{1}{2}+\frac{\sqrt3}{6})w\]

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0or maybe that is the wrong one

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@misty1212 because V = 2xW^2 , not as your equation.

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0\[(\frac{1}{2}\frac{\sqrt3}{6})w\] is probably better

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0hmm i get \[V=(w2x)(2w2x)x\] as a start

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0multiply out get \[V(x)=2 w^2 x6 w x^2+4 x^3\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@misty1212 not that, the w is the width of the box, 2w is the length of the box and x is its height

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0not the way you drew it for sure

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0oh i see then the way the question is asked, there is no maximum make w bigger and bigger the volume increases without bound

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@oldrin.bataku I saw you wrote something, where is it?? why don't you post it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for a fixed surface area? $$A=2w^2+2wx+2(2w)x\\\quad =2w^2+6wx$$so we have \(x=\frac{A2w^2}{6w}=\frac{A}{6w}\frac{w}3\) so $$V=2xw^2=2\left(\frac{A}{6w}\frac{w}3\right)w^2=\frac{Aw}{3}\frac{2w^3}{3}\\\frac{dV}{dw}=\frac{A}32w^2$$so $$\frac{A}32w^2=0\\2w^2=A/3\\w^2=A/6\\w=\sqrt{A/6}$$

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Where 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
 one year ago
Best ResponseYou've already chosen the best response.0plug it in: $$x=\frac{A2\cdot A/6}{6\cdot\sqrt{A/6}}=\frac{AA/3}{\sqrt{6A}}=\frac{2A/3}{\sqrt{6A}}=\frac{\sqrt{2A}}{3\sqrt3}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0I got exactly the same, but it is impossible when \(0<x<W/2\)