# Max f(x,y,z) = min{x, 5y+2z} subject to x+15y+7z=44

Max f(x,y,z) = min{x, 5y+2z} subject to x+15y+7z=44

As well, $x,y,z \geq 0$

I have guessed that the extrema point will be a point such that x=5y+2z and tried solving for the curve of intersection of z=(x-5y)/2 and the constraint x+15y+7z=44 and finding the max of the curve of intersection that results but is unfamiliar with the techniques involved in the ideas and is not sure whether that is the correct method for solving such a problem.

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$x+15y+7z=44 \Rightarrow x=44-(15y+7z)$

While $x,y,z \geq 0,\Rightarrow 15y+7z\leq 44$

$f(x,y,z)=min\{x,5y+2z\}=\begin{cases} 44-(15y+7z)& \text{ if } 44-(15y+7z)\leq 5y+2z \\ 5y+2z& \text{ if } 44-(15y+7z)\gt 5y+2z \end{cases}$

i.e.

$f(x,y,z)=\begin{cases} 44-(15y+7z)& \text{ if } 20y+9z\geq 44 \\ 5y+2z& \text{ if } 20y+9z\lt 44 \end{cases}$, with constraints $\begin{cases} y\geq 0\\z\geq 0\\15y+7z\leq 44\end{cases}$

This becomes a 2-d problem, mentioned as @Emmad Kareem. But it has 3-d coordinate, $(x=44-(15y+7z),y,z)$.

$Max\{f\}=11$ at $y=44/20, z=0$, namely, at the point $(11,11/5,0)$

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I am not sure where does the (Z) value of (0) come from if we only have 2 dimensional point? +1 for the details. – NoChance Nov 15 '11 at 9:25
@Emmad Kareem, it is a 2-d problem since x+15y+7z=44, which is a plane. The problem is reduced to a 2-d linear programming, but f change with 3-d point, (x,y,z). Am I make it clearer now? – puresky Nov 15 '11 at 14:33

This is more of a comment than a solution...

It is common to want to min. a function rather than a point.

I assume that the point of the form (x, 5y+2z) as in your question is not really a function.

Also, the point you have has no Z value, it should have been something like (a, b, c)

p=(44-15y-7z, 5y+2z) where $x,y,z \geq 0$