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 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 = ??
 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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asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3express 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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3yes  you can do it that way as well

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3now you need to find the value for y that makes dP/dy=0

chrislb22
 2 years ago
Best ResponseYou've already chosen the best response.1im stuck at that part

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3do you know how to use the product rule when differentiating?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3if not  this might help: http://en.wikipedia.org/wiki/Product_rule

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3or else expand the expression out and then differentiate

chrislb22
 2 years ago
Best ResponseYou've already chosen the best response.1can you help me with the first step

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3what method are you planning to use?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3product rule or expand and then differentiate?

chrislb22
 2 years ago
Best ResponseYou've already chosen the best response.1the easiest way to do it

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3do you know how to differentiate an expression like \(3x^2+4x^3+6x\)?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3ok  so first expand your expression:\[(12y)^2 (y)\]which is usually written in this form:\[y(12y)^2\]

chrislb22
 2 years ago
Best ResponseYou've already chosen the best response.1right i got that part, then what, find the derivative?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3yes  but can you please first expand that expression  what do you get?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3I 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.

chrislb22
 2 years ago
Best ResponseYou've already chosen the best response.1y(12y)(12y) = y(14424y+y^2)

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3correct so far  now distribute y into the braces

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3perfect  now differentiate with respect to y

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3again, 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)

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3*is to differentiate it and...

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3so you have now found that:\[P=144y24y^2+y^3\] you now need to calculate dP/dy

chrislb22
 2 years ago
Best ResponseYou've already chosen the best response.13(y^216y+48) .... is that correct?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3correct, now find the value(s) of y that will make this expression equal to zero

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3you should be able to factor the expression inside the braces

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3correct  now one of those y values will give the maximum value for P and the other will probably represent a minimum value for P

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3do you know how to tell whether you have found a max/min/point of inflection using the second derivative test?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3what 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.

chrislb22
 2 years ago
Best ResponseYou've already chosen the best response.1not 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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3well substituting y=12 into your expression (\(P=144y24y^2+y^3\)) 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).

chrislb22
 2 years ago
Best ResponseYou've already chosen the best response.1ok then how would i find Z?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3when you started the problem you established that: x = 12y and z= 12y so just substitute the two value for y into these expressions to get the corresponding two values each for x and z.

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3but 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.

chrislb22
 2 years ago
Best ResponseYou've already chosen the best response.1so i got 256 and 3456 as my answers

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3what do those values represent? how did you calculate them?

chrislb22
 2 years ago
Best ResponseYou've already chosen the best response.1y (4) = 256, y(12) = 3456, plugged both into 144y24y^2+ y^3

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3your calculation is not correct for y=12

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3use the factored form you originally got:\[P=y(12y)^2\]this should make it easier to calculate the values for P.

chrislb22
 2 years ago
Best ResponseYou've already chosen the best response.1wouldnt that equal to 0?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3so you have now found: y=12 gives P=0 y=4 gives P=256

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3so which one do you think is the maximum?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3correct  so problem solved. :)

chrislb22
 2 years ago
Best ResponseYou've already chosen the best response.1ok, can i ask you another question, a lot easier than this.. please?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3It 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.

chrislb22
 2 years ago
Best ResponseYou've already chosen the best response.1ok thanks for the help
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