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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.

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  1. Loser66
    • one year ago
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    |dw:1438823690584:dw|

  2. anonymous
    • one year ago
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    is the length 2 or width 2?

  3. Loser66
    • one year ago
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    no numbers there, just that

  4. anonymous
    • one year ago
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    2w = length of box?

  5. Loser66
    • one year ago
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    yes

  6. anonymous
    • one year ago
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    is the box a rectangle as stated or is it a square

  7. Loser66
    • one year ago
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    My attempt: Volume of the box is V = x*W*L = x*W (2W) = 2x W^2

  8. anonymous
    • one year ago
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    They said the length is 2x bigger, so w = 1 in that case

  9. Loser66
    • one year ago
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    Surface are of the open box is \(S = (L+2x)(W+2x) - 4x^2= (2W+2x)(W+2x)-4x^2\) \(S = 2W^2+4xW+2xW+4x^2-4x^2= 2W^2+6xW\)

  10. anonymous
    • one year ago
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    I gotta run, continue with someone else

  11. Mertsj
    • one year ago
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    How can you find the volume if you know nothing about the height of the box?

  12. Loser66
    • one year ago
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    Thanks

  13. anonymous
    • one year ago
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    @Mer I thought the same, anyways np loser.

  14. Loser66
    • one year ago
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    @Mertsj maximum of volume at the point of derivative of the Volume =0

  15. Loser66
    • one year ago
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    We consider Volume of the box is the function with respect to x and W

  16. Loser66
    • one year ago
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    \(V = 2xW^2\\V'= 2W^2 + 2xW*W'\)

  17. anonymous
    • one year ago
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    i think we're presuming maximum volume for a fixed perimeter, \(P=2(2x+w+2x+2w)\)

  18. anonymous
    • one year ago
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    so $$P=8x+6w\implies x=\frac18(P-6w)$$

  19. Loser66
    • one year ago
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    No, not that , the perimeter of what? if it is of the box, then you don't have x there, right?

  20. Loser66
    • one year ago
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    |dw:1438825047689:dw| that is perimeter. One more thing, the prof asked us to work with Surface area. I am sorry for this information

  21. anonymous
    • one year ago
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    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 $$\frac{dV}{dw}=0\\Pw-9w^2=0\\w(P-9w)=0\implies w=P/9$$

  22. Loser66
    • one year ago
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    oh, sorry, you are correct with perimeter. :)

  23. anonymous
    • one year ago
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    which gives $$V^*=\frac14(P-6\cdot P/9)(P/9)^2=\frac14(P/3)(P^2/81)=\frac{P^3}{4\cdot 243}$$

  24. Loser66
    • one year ago
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    I got what you meant. How about Surface area? how to work on it?

  25. misty1212
    • one year ago
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    i get \[v(x) = 2 w^2 x-6 w x^2+4 x^3\]does that help?

  26. anonymous
    • one year ago
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    and also \(x=\frac18(P-6w)=\frac18(P/3)=\frac{P}{24}\)

  27. Loser66
    • one year ago
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    @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

  28. misty1212
    • one year ago
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    yeah take the derivative wrt x get\[2 (w^2-6 w x+6 x^2)\]

  29. misty1212
    • one year ago
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    set equal to zero, solve for \(x\) get something like \[(\frac{1}{2}+\frac{\sqrt3}{6})w\]

  30. misty1212
    • one year ago
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    or maybe that is the wrong one

  31. Loser66
    • one year ago
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    @misty1212 because V = 2xW^2 , not as your equation.

  32. misty1212
    • one year ago
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    \[(\frac{1}{2}-\frac{\sqrt3}{6})w\] is probably better

  33. misty1212
    • one year ago
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    hmm i get \[V=(w-2x)(2w-2x)x\] as a start

  34. misty1212
    • one year ago
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    multiply out get \[V(x)=2 w^2 x-6 w x^2+4 x^3\]

  35. Loser66
    • one year ago
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    @misty1212 not that, the w is the width of the box, 2w is the length of the box and x is its height

  36. misty1212
    • one year ago
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    not the way you drew it for sure

  37. misty1212
    • one year ago
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    oh i see then the way the question is asked, there is no maximum make w bigger and bigger the volume increases without bound

  38. Loser66
    • one year ago
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    @oldrin.bataku I saw you wrote something, where is it?? why don't you post it?

  39. anonymous
    • one year ago
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    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 $$\frac{A}3-2w^2=0\\2w^2=A/3\\w^2=A/6\\w=\sqrt{A/6}$$

  40. Loser66
    • one year ago
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    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.

  41. anonymous
    • one year ago
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    plug it in: $$x=\frac{A-2\cdot A/6}{6\cdot\sqrt{A/6}}=\frac{A-A/3}{\sqrt{6A}}=\frac{2A/3}{\sqrt{6A}}=\frac{\sqrt{2A}}{3\sqrt3}$$

  42. anonymous
    • one year ago
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    so the ratio of w, x is just $$x/w=\frac{\sqrt{2A}\cdot\sqrt6}{3\sqrt3\cdot\sqrt{A}}=\frac{\sqrt2\cdot\sqrt2}{3}=\frac23$$

  43. Loser66
    • one year ago
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    I got exactly the same, but it is impossible when \(0<x<W/2\)

  44. Loser66
    • one year ago
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    |dw:1438827425494:dw|