Question

please help!!

Solve the optimization problem.
Maximize P = xyz with x + y = 12 and y + z = 12, and x, y, and z ≥ 0.

p = ??

Answer 1

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

Answer 2

x = 12-y and z= 12-y

Answer 3

(12-y)^2 (y)

Answer 4

yes - you can do it that way as well

Answer 5

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

Answer 6

im stuck at that part

Answer 7

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

Answer 8

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

Answer 9

or else expand the expression out and then differentiate

Answer 10

can you help me with the first step

Answer 11

what method are you planning to use?

Answer 12

product rule or expand and then differentiate?

Answer 13

the easiest way to do it

Answer 14

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

Answer 15

6(1+x+2x^2)

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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.

Answer 20

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

Answer 21

correct so far - now distribute y into the braces

Answer 22

144y-24y^2+y^3

Answer 23

perfect - now differentiate with respect to y

Answer 24

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)

Answer 25

*is to differentiate it and...

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

you should be able to factor the expression inside the braces

Answer 30

y= 4 , y= 12

Answer 31

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

Answer 32

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

Answer 33

no

Answer 34

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.

Answer 35

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

Answer 36

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).

Answer 37

ok then how would i find Z?

Answer 38

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.

Answer 39

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.

Answer 40

so i got 256 and 3456 as my answers

Answer 41

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

Answer 42

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

Answer 43

your calculation is not correct for y=12

Answer 44

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

Answer 45

wouldnt that equal to 0?

Answer 46

for y=12, yes

Answer 47

so you have now found:

y=12 gives P=0
y=4 gives P=256

Answer 48

so which one do you think is the maximum?

Answer 49

256

Answer 50

correct - so problem solved. :)

Answer 51

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

Answer 52

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.

Answer 53

ok thanks for the help

Answer 54

yw :)