10. Maximize z = 30x + 19y such that x + y ≤ 24, x + , y ≤ 16, x ≥ 0, y ≥ 0.

We are to maximize
z = 30x + 19y
subject to the constraints
 x + y ≤ 24,

 x ≥ 0, y ≥ 0.
Consider a set of rectangular cartesian axis OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of x + y = 24.
For x = 0, y = 24
For y = 0, x = 24
∴  line meets OX in A (24, 0) and OY in L(0, 24).
Again we draw the graph of 
For  
For  y = 0,  x = 16
∴ line meets OX in B (16, 0) and OY in M (0, 32).

Since feasible region is the region which satisfies all the constraints
∴ OBCL is the feasible region and O (0, 0), B (16, 0), C (8, 16), L (0, 24) are corner points.
At O(0, 0), z = 30(0) + 19 (0) = 0 + 0 = 0
At B(16, 0), z = 30 (16) + 19 (0) = 480 + 0 = 480
At C (8, 16), z = 30 (8) + 19 (16) = 240 + 304 = 544
At L (0, 24), z = 30 (0) + 19 (24) = 0 + 456 = 456
∴ maximum value = 544 at (8, 16).