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tmagloire1

  • one year ago

A box is to be constructed from a sheet of cardboard that is 20 cm by 50 cm by cutting out squares of length x by x from each corner and bending up the sides. What is the maximum volume this box could have? (Round your answer to two decimal places. Do not include units, for example, 10.22 cm would be 10.22.)

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  1. SolomonZelman
    • one year ago
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    calc I prob optimization, correct?

  2. SolomonZelman
    • one year ago
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    |dw:1437076558138:dw|

  3. tmagloire1
    • one year ago
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    Yep optimization

  4. tmagloire1
    • one year ago
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    I got 2029.95 but that isn't correct

  5. SolomonZelman
    • one year ago
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    |dw:1437076685341:dw|

  6. tmagloire1
    • one year ago
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    Yep that's how i set it up

  7. SolomonZelman
    • one year ago
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    do you know why I subtracted 2xfrom 20 and 50?

  8. SolomonZelman
    • one year ago
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    So you do know why we subtract 2x from 20 and 50, right/

  9. tmagloire1
    • one year ago
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    Yes i understand

  10. SolomonZelman
    • one year ago
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    So, now, we will create the function for volume. Find critical numbers Find the maximum of it (with limits we will later provide/make)

  11. SolomonZelman
    • one year ago
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    The volume would be height•lenght•width

  12. tmagloire1
    • one year ago
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    I got 4.468

  13. SolomonZelman
    • one year ago
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    and in this case v(x)=2x•(20-2x)•(50-2x)

  14. SolomonZelman
    • one year ago
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    where V stands for volume

  15. tmagloire1
    • one year ago
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    Why would it be 2x instead of just x

  16. SolomonZelman
    • one year ago
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    u need v`(x)

  17. SolomonZelman
    • one year ago
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    Why would it be 2x instead of just x, because you bend a cut-out of length x from both sides, don't you?

  18. SolomonZelman
    • one year ago
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    |dw:1437076945620:dw|

  19. tmagloire1
    • one year ago
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    ohh okay

  20. SolomonZelman
    • one year ago
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    so the side 20 and side 50 is decreased by 2x. Got that, good:)

  21. tmagloire1
    • one year ago
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    So now take the derivative of the function we got?

  22. SolomonZelman
    • one year ago
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    now we will do a regular process once we got the function V(x)=2x•(20-2x)•(50-2x)

  23. SolomonZelman
    • one year ago
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    I would perhaps simplify it a little instead of doing a product twice, but you can find the derivative as you wish:)

  24. SolomonZelman
    • one year ago
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    Oh, my bad

  25. SolomonZelman
    • one year ago
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    V(x)=x•(20-2x)•(50-2x) ^ ^ just x, not 2x.

  26. SolomonZelman
    • one year ago
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    but 20-2x and 50-2x are correct measures

  27. tmagloire1
    • one year ago
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    Okay so take the derivative of that then?

  28. SolomonZelman
    • one year ago
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    V(x)=x•(20-2x)•(50-2x) V(x)=x•(1000-40x-100x+4x²) V(x)=x•(1000-140x+4x²) I would do derivative, product now

  29. SolomonZelman
    • one year ago
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    V\(\rm ~'\)(x) = 1•(1000-140x+4x²)+x•(0-140+8x)

  30. tmagloire1
    • one year ago
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    So it will be 12x^2-280x-1000?

  31. SolomonZelman
    • one year ago
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    I would use wolfram, actually, because I am the laziest person on this planet.

  32. tmagloire1
    • one year ago
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    Hahaha i just used Wolfram for it because i didn't feel like taking the derivative myself xD

  33. SolomonZelman
    • one year ago
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    V'(x)=4(3x²-70x+250) by wolfram.... I did it like this because wolfram can give a handy version faster

  34. SolomonZelman
    • one year ago
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    set V'(x)=0 to find your critical number

  35. SolomonZelman
    • one year ago
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    numbers*

  36. SolomonZelman
    • one year ago
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    remember tho, that 0>x>10 or else your volume is going to be =0 or negative.

  37. tmagloire1
    • one year ago
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    I got 11.66 :/

  38. SolomonZelman
    • one year ago
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    nope

  39. SolomonZelman
    • one year ago
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    that wouldn't make sense, because you get a negative volume if you plug it in our volume function.

  40. SolomonZelman
    • one year ago
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    As again 0<x<10

  41. SolomonZelman
    • one year ago
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    find the roots of 4(3x²-70x+250)=0

  42. tmagloire1
    • one year ago
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    26.48 and -3.146 which still doesn't make sense right?

  43. SolomonZelman
    • one year ago
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    And I will tell you will get the ±, and only take the minu one, because the one with the + is greater than 10

  44. SolomonZelman
    • one year ago
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    what crtical number(s) did you get?

  45. tmagloire1
    • one year ago
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    I am getting -60 now :// sos

  46. SolomonZelman
    • one year ago
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    |dw:1437077863587:dw|

  47. SolomonZelman
    • one year ago
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    take the -, because + will get x>10

  48. SolomonZelman
    • one year ago
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    35-5√19 ≈ 4.4

  49. tmagloire1
    • one year ago
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    Oh I must have been doing some crazy math error when taking the crit #

  50. tmagloire1
    • one year ago
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    so take 4.4 and use it in the equation at the beginning?

  51. SolomonZelman
    • one year ago
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    yes,

  52. tmagloire1
    • one year ago
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    i got -2000 ._.

  53. SolomonZelman
    • one year ago
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    you did something wrong again-:(

  54. tmagloire1
    • one year ago
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    2030.3*

  55. SolomonZelman
    • one year ago
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    Yup, that is right

  56. SolomonZelman
    • one year ago
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    2030.336

  57. tmagloire1
    • one year ago
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    Oh okay! Thank you so much of all the help. I think I understand it now!!

  58. SolomonZelman
    • one year ago
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    Yes.... overview. 1. We created a function V(x)=x • (20-2x) • (50-2x) 2. Constraint x>0, because otherwise volume is negative or 0 (based on the first, x, component of V(x)). x<10, because otherwise (based on the 20-2x part), the volume would be 0 or negative. 3. We differentiated the function to get 2 critical numbers (that is two solutions of V'(x)=0) NOTE: Critical numbers (usually) are also values for which v'(x) is undefined, if defined for initial function v(x) - (but no such values in this case). Also, critical numbers usually can be closed interval boundaries, but our interval is open (we didn't include 0 or 10), so that is not our case. So in our case only solutions to v`(x)=0 are critical numbers. 4. We excluded the (35+5√19)/3 solution based on our constraint (because it is greater than 10), and rather used the only solution remaining, i.e. (35-5√19)/3. 5. We got the answr

  59. SolomonZelman
    • one year ago
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    yw

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