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

- anonymous

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- asnaseer

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

- anonymous

x = 12-y and z= 12-y

- anonymous

(12-y)^2 (y)

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## More answers

- asnaseer

yes - you can do it that way as well

- asnaseer

now you need to find the value for y that makes dP/dy=0

- anonymous

im stuck at that part

- asnaseer

do you know how to use the product rule when differentiating?

- asnaseer

if not - this might help: http://en.wikipedia.org/wiki/Product_rule

- asnaseer

or else expand the expression out and then differentiate

- anonymous

can you help me with the first step

- asnaseer

what method are you planning to use?

- asnaseer

product rule or expand and then differentiate?

- anonymous

the easiest way to do it

- asnaseer

do you know how to differentiate an expression like \(3x^2+4x^3+6x\)?

- anonymous

6(1+x+2x^2)

- asnaseer

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

- anonymous

right i got that part, then what, find the derivative?

- asnaseer

yes - but can you please first expand that expression - what do you get?

- asnaseer

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.

- anonymous

y(12-y)(12-y) = y(144-24y+y^2)

- asnaseer

correct so far - now distribute y into the braces

- anonymous

144y-24y^2+y^3

- asnaseer

perfect - now differentiate with respect to y

- asnaseer

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)

- asnaseer

*is to differentiate it and...

- asnaseer

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

- anonymous

3(y^2-16y+48) .... is that correct?

- asnaseer

correct, now find the value(s) of y that will make this expression equal to zero

- asnaseer

you should be able to factor the expression inside the braces

- anonymous

y= 4 , y= 12

- asnaseer

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

- asnaseer

do you know how to tell whether you have found a max/min/point of inflection using the second derivative test?

- anonymous

no

- asnaseer

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.

- anonymous

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

- asnaseer

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).

- anonymous

ok then how would i find Z?

- asnaseer

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.

- asnaseer

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.

- anonymous

so i got 256 and 3456 as my answers

- asnaseer

what do those values represent? how did you calculate them?

- anonymous

y (4) = 256, y(12) = 3456, plugged both into 144y-24y^2+ y^3

- asnaseer

your calculation is not correct for y=12

- asnaseer

use the factored form you originally got:\[P=y(12-y)^2\]this should make it easier to calculate the values for P.

- anonymous

wouldnt that equal to 0?

- asnaseer

for y=12, yes

- asnaseer

so you have now found:
y=12 gives P=0
y=4 gives P=256

- asnaseer

so which one do you think is the maximum?

- anonymous

256

- asnaseer

correct - so problem solved. :)

- anonymous

ok, can i ask you another question, a lot easier than this.. please?

- asnaseer

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.

- anonymous

ok thanks for the help

- asnaseer

yw :)

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