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.
 A: $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 $(11,11/5,0)$
A: 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)
Using your constraint, x=44-15y-7z
so you the point becomes:
p=(44-15y-7z, 5y+2z) where $x,y,z \geq 0$
but the above point is in 2-d plane since it has no 3rd coordinate.
I am not sure how to proceed further...
