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x = 12-y and z= 12-y

(12-y)^2 (y)

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)

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

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

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

*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

no

ok then how would i find 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

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?

ok thanks for the help

yw :)