The easiest way to visualise and solve linear programming problems is to sketch graphs. Replace the inequality signs of the constraints ("subject to") with equal signs and plot the lines. Less than or equal to means the area to the left of the line, while greater than or equal to means the area to the right. If these areas are shaded, the feasibility region is the area in which all the shadings coincide. When sketching, mark x and y intercepts (these usually apply because x,y≥0 are usually included in the constraints) and mark where lines intersect one another. Some of these are likely to create the vertices of the feasibility region. Sketches don't need to be to scale but relative positioning of intercepts and intersections are important especially when some constraints are very similar. Make sure that you take notice whether it's the maximum or minimum value that needs to be measured. Evaluate the required quantity at each of the vertices and you'll see which vertex corresponds to the max or min. It's easy to make a mistake in these problems, so when you think you've found the solution, go back and check that the solution satisfies all the constraints.
(1)
The first two constraints have the same x intercept (200,0) but the y intercept of the first is outside of that of the second, so one vertex of the feasible region is the second y intercept 560/0.4=1400, giving the coordinates (0,1400).
The next vertex is the x intercept at 200 (at (200,0)).
The x and y intercepts of the third constraint give us (0,3.9/0.03)=(0,130) and (3.9/0.04,0)=(97.5,0).
We can now evaluate p=25x+20y for each of the four vertices:
(0,1400): 28,000;
(200,0): 5,000;
(0,130): 2,600;
(97.5,0): 2,437.5.
Max p is therefore x=0, y=1400.
(2)
2 vertices are: (0,300), (150,0). (13,0), (0,130) are outside the feasibility region because they don't satisfy the other constraint. These are x or y intercepts in this case. To find the x intercept set y=0 in the inequality (treated as an equation) and solve for x. Vice versa for finding the y intercept.
1.8x+0.9y≥270. (y intercept (x=0)=270/0.9=300, x intercept (y=0)=270/1.8=150), area to the right;
0.3x+0.03y≥3.9. (x intercept=3.9/0.3=13, y intercept=3.9/0.03=130), part of the area to the right is not within the feasibility region;
Plug each of these into c=30x+10y:
(0,300): 3,000;
(150,0): 4,500;
(13,0): 390; (not in feasibility region)
(0,130): 1,300. (not in feasibility region.
Minimum c is x=0, y=3000.
