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chrislb22

  • 2 years ago

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

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  1. asnaseer
    • 2 years ago
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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

  2. chrislb22
    • 2 years ago
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    x = 12-y and z= 12-y

  3. chrislb22
    • 2 years ago
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    (12-y)^2 (y)

  4. asnaseer
    • 2 years ago
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    yes - you can do it that way as well

  5. asnaseer
    • 2 years ago
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    now you need to find the value for y that makes dP/dy=0

  6. chrislb22
    • 2 years ago
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    im stuck at that part

  7. asnaseer
    • 2 years ago
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    do you know how to use the product rule when differentiating?

  8. asnaseer
    • 2 years ago
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    if not - this might help: http://en.wikipedia.org/wiki/Product_rule

  9. asnaseer
    • 2 years ago
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    or else expand the expression out and then differentiate

  10. chrislb22
    • 2 years ago
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    can you help me with the first step

  11. asnaseer
    • 2 years ago
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    what method are you planning to use?

  12. asnaseer
    • 2 years ago
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    product rule or expand and then differentiate?

  13. chrislb22
    • 2 years ago
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    the easiest way to do it

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

  15. chrislb22
    • 2 years ago
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    6(1+x+2x^2)

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

  17. chrislb22
    • 2 years ago
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    right i got that part, then what, find the derivative?

  18. asnaseer
    • 2 years ago
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    yes - but can you please first expand that expression - what do you get?

  19. asnaseer
    • 2 years ago
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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.

  20. chrislb22
    • 2 years ago
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    y(12-y)(12-y) = y(144-24y+y^2)

  21. asnaseer
    • 2 years ago
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    correct so far - now distribute y into the braces

  22. chrislb22
    • 2 years ago
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    144y-24y^2+y^3

  23. asnaseer
    • 2 years ago
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    perfect - now differentiate with respect to y

  24. asnaseer
    • 2 years ago
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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)

  25. asnaseer
    • 2 years ago
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    *is to differentiate it and...

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

  27. chrislb22
    • 2 years ago
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    3(y^2-16y+48) .... is that correct?

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

  29. asnaseer
    • 2 years ago
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    you should be able to factor the expression inside the braces

  30. chrislb22
    • 2 years ago
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    y= 4 , y= 12

  31. asnaseer
    • 2 years ago
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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

  32. asnaseer
    • 2 years ago
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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?

  33. chrislb22
    • 2 years ago
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    no

  34. asnaseer
    • 2 years ago
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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.

  35. chrislb22
    • 2 years ago
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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

  36. asnaseer
    • 2 years ago
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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).

  37. chrislb22
    • 2 years ago
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    ok then how would i find Z?

  38. asnaseer
    • 2 years ago
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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.

  39. asnaseer
    • 2 years ago
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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.

  40. chrislb22
    • 2 years ago
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    so i got 256 and 3456 as my answers

  41. asnaseer
    • 2 years ago
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    what do those values represent? how did you calculate them?

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

  43. asnaseer
    • 2 years ago
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    your calculation is not correct for y=12

  44. asnaseer
    • 2 years ago
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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.

  45. chrislb22
    • 2 years ago
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    wouldnt that equal to 0?

  46. asnaseer
    • 2 years ago
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    for y=12, yes

  47. asnaseer
    • 2 years ago
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    so you have now found: y=12 gives P=0 y=4 gives P=256

  48. asnaseer
    • 2 years ago
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    so which one do you think is the maximum?

  49. chrislb22
    • 2 years ago
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    256

  50. asnaseer
    • 2 years ago
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    correct - so problem solved. :)

  51. chrislb22
    • 2 years ago
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    ok, can i ask you another question, a lot easier than this.. please?

  52. asnaseer
    • 2 years ago
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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.

  53. chrislb22
    • 2 years ago
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    ok thanks for the help

  54. asnaseer
    • 2 years ago
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    yw :)

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