Minimum value of $6x+7.5y$ What is the minimum value of $6x+7.5y\,$given that$\,x+y\geq 48 , x\geq 12 , y\geq 8\,$and$\,x,y\,$are positive integers
Since$\,x\geq 12 \implies 6x\geq 72\,$and$\\y\geq 8 \implies7.5y\geq 60$
$6x+7.5y\geq 132\,$
Do I need to solve the system
$\,x+y\geq 48$ 
$6x+7.5y\geq 132\,$?
 A: $6x+7.5y=6(x+y)+1.5y\ge48\cdot6+1.5\cdot8=300$
A: It is called the integer linear programming problem.
Systematic way of solving it is to draw the feasible region and check the corner points in the feasible region.
Refer to the graph:

The feasible region is on the right of blue line ($x\ge 12$), above the green line ($y\ge 8$) and above the red line ($x+y\ge 48$). The black dotted line is the contour line of the objective function ($6x+7.5y$). Hence, the minimum value ($300$) of the objective function occurs at the point $(40,8)$.
Note: The corner point may be non-integers, then you must check the points with integer coordinates in the feasible region closest to the corner points.
A: We may consider the function $f(x,y)= 6x+7.5y$. It is obvious that the minimum, if there is, should be in the compact domain $$\,x+y\geq 48 , 100\geq{x}\geq 12 , 100\geq{y}\geq 8\,$$. But $f$ is continous so there is a minimum there and as $\nabla{f}$ is $(6, 7.5)$ i.e not zero,the minimum should be on the boundary of that domain and not in its interior. We can check now that on the lines $y=8$ , $x=12$ , $x+y=48$, $y=100$, and $x=100$,   the point $(40,8)$ gives us the minimum.
A: Just to put it yet another way.  (Inefficient--  If you want efficiency go with abc's excellent answer-- but hopefully a tiny bit insightful.)
Imagine the line $x = 12$. (This is vertical line.) Any acceptable point must lie to the right of that.  
Imagine the line $y = 8$.  (This is a horizontal line.) Any acceptable point must lie above that.
Imagine the line $x + y = 48$.  (This is a line with a slope of $-1$ and a $y$-intercept at $(0,48)$ and an $x$-intercept at $(48,0$.)  Any acceptable point must lie to the right and above that.
This forms an acceptable area that resembles a truncated quadrant.  Anything where $x \ge 48$ and $y\ge 8$ or $y \ge 48; x \ge 12$ is good.  Nothing where either $x < 12$ or $y < 12$ is good.  And for the rectangle where $12\le x \le 48$ and $8\le y \le 48$ .... The line $x + y = 48$ cuts through that (it could have been that the line missed it entirely and we would have to worry about it, but that wasn't the case) and we can pick any point in rectangle above and to the right of that line.
We'll call this "The Area".
Now consider a line $6x + 7.5y = k$ for some unknown $k$.  This will be a  line with a slope of $-\frac 6{7.5} = - \frac 45$, but we do not know what the $x$ and $y$ intersepts are.  We want to find the least possible value of $k$ where the line pass through The Area.
Different values of $k$ will produce different lines but they will all have slope $-\frac 45$ and be parallel to each other.
Bear with me.
A small value of $k$ will produce a line that goes "too low" and misses The Area entirely.  A high value of $k$ will slice through The Area and have an infinite number of points in The Area are below and to the left of the line (and for those $(x,y)$ points we'd have  $6x + 7.5y< k$).
So we want to find a $k$ that produces a parallel line that just touches the area.
Please continue to bear with me.
The slope of the parallel lines is $-\frac 45$ and the slopes of the borders of The Area are $-1$ and $0$ and $\infty$.  So a parallel line that touches The Area just once will be in one of the two corners.  Either $(12, y)= (1, 48-12 ) = (12,36)$ or $(x,8) = (48-8, 8) = (40,8)$.
Now the slope of $-\frac 45$ is "shallower" than the slope of $-1$ so the line we want most pass through the lower corner, $(40,8)$.  (The line passing through the higher corner, $(12,36)$, will go into The Area because the slope in shallower than The Area's border edge; But the line passing through $(40,8)$ hits the area just once.  Draw a picture to see this.  farruhota includes a graph.  If you shade it I hope you can see the shapes and The Area that I'm talking about.)
(Added my one picture:  It's a little busy (very busy) but I hope it is clear.  Our goal is to find a purple parallel line that hits The Area at exactly one point.)

So we want the line $6x + 7.5y = k$ which goes through $(40,8)$.
In other word $6*40 + 7.5*8 = k$ or $k = 240 + 60 = 300$. 
The lowest possible value is $6x + 7.5y = 300$.  
Just to beat this horse further.  A line with slope $-\frac 45$ passing through $(12,36)$ will have a $k$ value of $6*12 + 7.5*36 = 72 + 360 = 432$.  A line $6x + 7.5y = k$ for $300 < k < 432$ will enter The Area somewhere on the sloped edge.  
If $k > 432$ then the line $6x + 7.5y = k$ will enter The Area at $(12, y)$ (through the vertical edge) where $y > 36$.  And if $k < 300$ the line will never enter the area at all because at either $(12, y)$ we'd have $y < 8$ or at $(x, 8)$ we'd have $x < 12$.
