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chrislb22
Group Title
Solve the optimization problem
Maximize P = xyz with x + y = 36 and y + z = 36, and x, y, and z ≥ 0
 one year ago
 one year ago
chrislb22 Group Title
Solve the optimization problem Maximize P = xyz with x + y = 36 and y + z = 36, and x, y, and z ≥ 0
 one year ago
 one year ago

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LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Well, we know: \[ P=xyz, \\x+y=36,\\y+z=36 \]So: \[ x=36y \]And: \[ z=36y \]Therefore: \[ P=(36y)^2y \]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.
 one year ago

chrislb22 Group TitleBest ResponseYou've already chosen the best response.0
@lolwolf where did you get y=12?
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

chrislb22 Group TitleBest ResponseYou've already chosen the best response.0
i understand that, but just wondering where you got the 12 from
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Well, personally, I did a derivative test: \[ \frac{d}{dy}\left((36y)^2y\right)=36^2144y+3y^2=0 \]Solving this we get: \[y\in\{12,36\} \] But, 36 would make no sense as \(P=0\), so, therefore \(y=12\).
 one year ago
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