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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
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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asnaseer Group TitleBest ResponseYou've already chosen the best response.3
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

chrislb22 Group TitleBest ResponseYou've already chosen the best response.1
x = 12y and z= 12y
 one year ago

chrislb22 Group TitleBest ResponseYou've already chosen the best response.1
(12y)^2 (y)
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
yes  you can do it that way as well
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
now you need to find the value for y that makes dP/dy=0
 one year ago

chrislb22 Group TitleBest ResponseYou've already chosen the best response.1
im stuck at that part
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
do you know how to use the product rule when differentiating?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
if not  this might help: http://en.wikipedia.org/wiki/Product_rule
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
or else expand the expression out and then differentiate
 one year ago

chrislb22 Group TitleBest ResponseYou've already chosen the best response.1
can you help me with the first step
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
what method are you planning to use?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
product rule or expand and then differentiate?
 one year ago

chrislb22 Group TitleBest ResponseYou've already chosen the best response.1
the easiest way to do it
 one year ago

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

chrislb22 Group TitleBest ResponseYou've already chosen the best response.1
6(1+x+2x^2)
 one year ago

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

chrislb22 Group TitleBest ResponseYou've already chosen the best response.1
right i got that part, then what, find the derivative?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
yes  but can you please first expand that expression  what do you get?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
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

chrislb22 Group TitleBest ResponseYou've already chosen the best response.1
y(12y)(12y) = y(14424y+y^2)
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
correct so far  now distribute y into the braces
 one year ago

chrislb22 Group TitleBest ResponseYou've already chosen the best response.1
144y24y^2+y^3
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
perfect  now differentiate with respect to y
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
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

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
*is to differentiate it and...
 one year ago

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

chrislb22 Group TitleBest ResponseYou've already chosen the best response.1
3(y^216y+48) .... is that correct?
 one year ago

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

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
you should be able to factor the expression inside the braces
 one year ago

chrislb22 Group TitleBest ResponseYou've already chosen the best response.1
y= 4 , y= 12
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
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

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

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
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

chrislb22 Group TitleBest ResponseYou've already chosen the best response.1
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

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
well 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).
 one year ago

chrislb22 Group TitleBest ResponseYou've already chosen the best response.1
ok then how would i find Z?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
when 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.
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
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

chrislb22 Group TitleBest ResponseYou've already chosen the best response.1
so i got 256 and 3456 as my answers
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
what do those values represent? how did you calculate them?
 one year ago

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

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
your calculation is not correct for y=12
 one year ago

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

chrislb22 Group TitleBest ResponseYou've already chosen the best response.1
wouldnt that equal to 0?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
for y=12, yes
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
so you have now found: y=12 gives P=0 y=4 gives P=256
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
so which one do you think is the maximum?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
correct  so problem solved. :)
 one year ago

chrislb22 Group TitleBest ResponseYou've already chosen the best response.1
ok, can i ask you another question, a lot easier than this.. please?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
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

chrislb22 Group TitleBest ResponseYou've already chosen the best response.1
ok thanks for the help
 one year ago
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