Maximize P=xyz with x+y=30, y+z=30, and x,y, and z >= 0.(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

\begin{array}{l}

{

\\

y = 30 - x \\

\\

z = 30 - y\,\, \Rightarrow \,\,z = 30 - 30 - x = - x \\

\end{array}

[/tex]

This is a problem, because if z is negative x according to the question, they must both be 0 or positive. This isn't a problem if x = 0, and that's what I get as 1 of two solutions by differentiating P when written in terms of x, and setting it to 0.

[tex]

\begin{array}{l}

P = x(30 - x)( - x) = \left( { - x} \right)^2 (30 - x) = - 30x^2 + x^3 \\

P = x^3 - 30x^2 \\

\\

P' = 3x^2 - 60x \\

P' = x(3x - 60) \\

\end{array}

[/tex]

[tex]

x = 0

[/tex]

and

[tex]

3x - 60 = 0\,\, \Rightarrow \,\,3x = 60\,\, \Rightarrow \,\,x = \frac{{60}}{3} = 20

[/tex]

The z=-x isn't a problem in the x=0 solution, but it is a problem in the x=20 solution. This gives me x=20, y=10, z=-20 P=xyz=-4000.

The back of the book gives x=20, y=10, z=20 P=4000.

If z >= 0 and I get -20 for z, I can't just switch it to positive, can I? Don't I have to discard that solution?

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# Homework Help: Help with Maximize

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