## anonymous 3 years ago Solve the optimization problem Maximize P = xyz with x + y = 36 and y + z = 36, and x, y, and z ≥ 0

1. anonymous

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.

2. anonymous

@lolwolf where did you get y=12?

3. anonymous

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.

4. anonymous

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

5. anonymous

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 $$y=12$$.

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