Question

Solve the optimization problem

Maximize P = xyz with x + y = 36 and y + z = 36, and x, y, and z ≥ 0

Answer 1

Well, we know:
\[
P=xyz, \\x+y=36,\\y+z=36
\]So:
\[
x=36-y
\]And:
\[
z=36-y
\]Therefore:
\[
P=(36-y)^2y
\]Find a maximum value for this case. I don't know how your teacher wants you to do this, but:
At \(y=12\), we have:
\[
P\big|_{y=12}=6912
\]Which is a maximum, for when, \(x, y, z\ge0\)
It should be relatively simple to solve for the other values.

Answer 2

@lolwolf where did you get y=12?

Answer 3

You can do two things:
1) Graph it, and then find the maximum, or
2) You can take the derivative and, using a candidate test, get the maximum.

Answer 4

i understand that, but just wondering where you got the 12 from

Answer 5

Well, personally, I did a derivative test:
\[
\frac{d}{dy}\left((36-y)^2y\right)=36^2-144y+3y^2=0
\]Solving this we get:
\[y\in\{12,36\}
\]
But, 36 would make no sense as \(P=0\), so, therefore \(y=12\).