$Minimize$ $z=-2x-5y$ subject to $3x+4y\ge 5$ , $x\ge 0$ , $y\ge 0$. Consider the linear programming problem:
$Minimize$ $z=-2x-5y$ subject to $3x+4y\ge 5$ , $x\ge 0$ , $y\ge 0$.
Which is correct ?
(A) Set of feasible solutions is empty.
(B) Set of feasible solution is non-empty but there are no optimal solution.
(C) Optimal value is attained at $(0,5/4)$.
(D) Optimal value is attained at $(5/3,0)$.
I have no idea about this problem..I tried to draw the figure but I did not get any region , as there are only one constraint is given..I could not deal with only one constraint..How I solve this?
 A: buy some graph paper. It has little squares. One thing that has been confirmed on MSE is that a student who relies on software to draw diagrams will never learn to visualize, neither in two dimensions nor three. It is, of course, slower to do it yourself. 
Next, evaluate your objective function $-2x-5y$ at some integer points in the region, $(1,0), $ $(0,1), $ $(2,0), $ $(1,1), $ $(0,2), $ 

A: Consider a 2-d surface. $x\ge 0$ means only points to the left of the origin, $y\ge 0$ means only points to the upwards of origin. So, these two conditions together mean only points in the first quadrant. Now draw the line $3x+4y=5$. This line divides the first quadrant (and the entire space) into two regions, we want to know $3x+4y\ge 5$ refers to which region. Pick any point say (2,2). Clearly, this satisfies $3x+4y\ge 5$, so this inequality refers to the region where this point lies. Now you have a region bounded by three constraints. Now just find out where does the line $z=-2x-5y$ get minimised in this region.
