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chrislb22 Group Title

please help!! Solve the optimization problem. Maximize P = xyz with x + y = 12 and y + z = 12, and x, y, and z ≥ 0. p = ??

  • one year ago
  • one year ago

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  1. asnaseer Group Title
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    express P in terms of x only then find the value of x that makes dP/dx=0 use this value to find the max value for P

    • one year ago
  2. chrislb22 Group Title
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    x = 12-y and z= 12-y

    • one year ago
  3. chrislb22 Group Title
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    (12-y)^2 (y)

    • one year ago
  4. asnaseer Group Title
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    yes - you can do it that way as well

    • one year ago
  5. asnaseer Group Title
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    now you need to find the value for y that makes dP/dy=0

    • one year ago
  6. chrislb22 Group Title
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    im stuck at that part

    • one year ago
  7. asnaseer Group Title
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    do you know how to use the product rule when differentiating?

    • one year ago
  8. asnaseer Group Title
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    if not - this might help: http://en.wikipedia.org/wiki/Product_rule

    • one year ago
  9. asnaseer Group Title
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    or else expand the expression out and then differentiate

    • one year ago
  10. chrislb22 Group Title
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    can you help me with the first step

    • one year ago
  11. asnaseer Group Title
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    what method are you planning to use?

    • one year ago
  12. asnaseer Group Title
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    product rule or expand and then differentiate?

    • one year ago
  13. chrislb22 Group Title
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    the easiest way to do it

    • one year ago
  14. asnaseer Group Title
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    do you know how to differentiate an expression like \(3x^2+4x^3+6x\)?

    • one year ago
  15. chrislb22 Group Title
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    6(1+x+2x^2)

    • one year ago
  16. asnaseer Group Title
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    ok - so first expand your expression:\[(12-y)^2 (y)\]which is usually written in this form:\[y(12-y)^2\]

    • one year ago
  17. chrislb22 Group Title
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    right i got that part, then what, find the derivative?

    • one year ago
  18. asnaseer Group Title
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    yes - but can you please first expand that expression - what do you get?

    • one year ago
  19. asnaseer Group Title
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    I am /assuming/ you do not know how to use the product or chain rule. which is why I first asked if you at least knew how to differentiate a general polynomial expression. since you do, then for you, the easiest would be to first expand that expression to get a polynomial expression and then differentiate it.

    • one year ago
  20. chrislb22 Group Title
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    y(12-y)(12-y) = y(144-24y+y^2)

    • one year ago
  21. asnaseer Group Title
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    correct so far - now distribute y into the braces

    • one year ago
  22. chrislb22 Group Title
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    144y-24y^2+y^3

    • one year ago
  23. asnaseer Group Title
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    perfect - now differentiate with respect to y

    • one year ago
  24. asnaseer Group Title
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    again, here I am /assuming/ that you are aware that to find the max/min of a polynomial expression (if one exists) if to differentiate it and set the resulting expression to zero. that will give you the value for the unknown for which the original expression has a max/min (or point of inflection)

    • one year ago
  25. asnaseer Group Title
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    *is to differentiate it and...

    • one year ago
  26. asnaseer Group Title
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    so you have now found that:\[P=144y-24y^2+y^3\] you now need to calculate dP/dy

    • one year ago
  27. chrislb22 Group Title
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    3(y^2-16y+48) .... is that correct?

    • one year ago
  28. asnaseer Group Title
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    correct, now find the value(s) of y that will make this expression equal to zero

    • one year ago
  29. asnaseer Group Title
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    you should be able to factor the expression inside the braces

    • one year ago
  30. chrislb22 Group Title
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    y= 4 , y= 12

    • one year ago
  31. asnaseer Group Title
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    correct - now one of those y values will give the maximum value for P and the other will probably represent a minimum value for P

    • one year ago
  32. asnaseer Group Title
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    do you know how to tell whether you have found a max/min/point of inflection using the second derivative test?

    • one year ago
  33. chrislb22 Group Title
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    no

    • one year ago
  34. asnaseer Group Title
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    what have you been taught about local maximums and minimums? I just want to get a idea of what approach to show you that best matches what you have been taught.

    • one year ago
  35. chrislb22 Group Title
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    not much, i actually have an assignment due very soon so i was just hoping what the next step would be after finding y=12, y=4

    • one year ago
  36. asnaseer Group Title
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    well substituting y=12 into your expression (\(P=144y-24y^2+y^3\)) will give you one value for P. And substituting y=4 into this expression will give you another value for P. One of these will represent the maximum value that P can have and the other will represent the minimum value that P can have (given the original constraints in the problem).

    • one year ago
  37. chrislb22 Group Title
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    ok then how would i find Z?

    • one year ago
  38. asnaseer Group Title
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    when you started the problem you established that: x = 12-y and z= 12-y so just substitute the two value for y into these expressions to get the corresponding two values each for x and z.

    • one year ago
  39. asnaseer Group Title
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    but first you need to work out which of the two y values gives you the maximum P value. Then use just that vaue of y to get x and z.

    • one year ago
  40. chrislb22 Group Title
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    so i got 256 and 3456 as my answers

    • one year ago
  41. asnaseer Group Title
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    what do those values represent? how did you calculate them?

    • one year ago
  42. chrislb22 Group Title
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    y (4) = 256, y(12) = 3456, plugged both into 144y-24y^2+ y^3

    • one year ago
  43. asnaseer Group Title
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    your calculation is not correct for y=12

    • one year ago
  44. asnaseer Group Title
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    use the factored form you originally got:\[P=y(12-y)^2\]this should make it easier to calculate the values for P.

    • one year ago
  45. chrislb22 Group Title
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    wouldnt that equal to 0?

    • one year ago
  46. asnaseer Group Title
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    for y=12, yes

    • one year ago
  47. asnaseer Group Title
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    so you have now found: y=12 gives P=0 y=4 gives P=256

    • one year ago
  48. asnaseer Group Title
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    so which one do you think is the maximum?

    • one year ago
  49. chrislb22 Group Title
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    256

    • one year ago
  50. asnaseer Group Title
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    correct - so problem solved. :)

    • one year ago
  51. chrislb22 Group Title
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    ok, can i ask you another question, a lot easier than this.. please?

    • one year ago
  52. asnaseer Group Title
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    It is very late where I am and I need to catch some sleep so that I can get to work tomorrow fully awake. I suggest you post your new question in the list on the left - there are plenty of very good people on this site who will surely come to your aid.

    • one year ago
  53. chrislb22 Group Title
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    ok thanks for the help

    • one year ago
  54. asnaseer Group Title
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    yw :)

    • one year ago
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