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

Mathematics
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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
x = 12-y and z= 12-y
(12-y)^2 (y)

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Other answers:

yes - you can do it that way as well
now you need to find the value for y that makes dP/dy=0
im stuck at that part
do you know how to use the product rule when differentiating?
if not - this might help: http://en.wikipedia.org/wiki/Product_rule
or else expand the expression out and then differentiate
can you help me with the first step
what method are you planning to use?
product rule or expand and then differentiate?
the easiest way to do it
do you know how to differentiate an expression like \(3x^2+4x^3+6x\)?
6(1+x+2x^2)
ok - so first expand your expression:\[(12-y)^2 (y)\]which is usually written in this form:\[y(12-y)^2\]
right i got that part, then what, find the derivative?
yes - but can you please first expand that expression - what do you get?
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.
y(12-y)(12-y) = y(144-24y+y^2)
correct so far - now distribute y into the braces
144y-24y^2+y^3
perfect - now differentiate with respect to y
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)
*is to differentiate it and...
so you have now found that:\[P=144y-24y^2+y^3\] you now need to calculate dP/dy
3(y^2-16y+48) .... is that correct?
correct, now find the value(s) of y that will make this expression equal to zero
you should be able to factor the expression inside the braces
y= 4 , y= 12
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
do you know how to tell whether you have found a max/min/point of inflection using the second derivative test?
no
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.
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
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).
ok then how would i find Z?
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.
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.
so i got 256 and 3456 as my answers
what do those values represent? how did you calculate them?
y (4) = 256, y(12) = 3456, plugged both into 144y-24y^2+ y^3
your calculation is not correct for y=12
use the factored form you originally got:\[P=y(12-y)^2\]this should make it easier to calculate the values for P.
wouldnt that equal to 0?
for y=12, yes
so you have now found: y=12 gives P=0 y=4 gives P=256
so which one do you think is the maximum?
256
correct - so problem solved. :)
ok, can i ask you another question, a lot easier than this.. please?
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.
ok thanks for the help
yw :)

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