Use graphical methods to solve each of the following linear programming problems.?

Question

A. Minimize Z=X+3y
subject to the constraints x+y < or equal to 10
5x+2y > or equal to 20
-x+2y > or equal to 0
x > or equal to 0
y > or equal to 0

B. Maximize Z=2x+2y Subject to the constraints:
x-y < or equal to 10
5x+3y < or equal to 75
x+y < or equal to 20
x> or equal to 0
y> or equal to 0

Answer 1

The basic idea is:
draw a set of axes
get all your constraints into y=mx+b form
draw all 3 lines in the range specified and shade the appropriate side
your answer will be staring at you.

for the first one:
x + y < 10 becomes y = -x + 10 you draw this line and shade everything to the left between the line and the axes (you stop at the axes because you have constraints x, y > 0)
5x + 2y > 20 becomes y = -5/2x + 10. shade to the right now (because it's a > symbol)
-x + 2y > 0 becomes y = 1/2x. shade to the right again

you now will have a region that was shaded from all 3 steps. in a perfect world, this region would be a point, and that would be your answer. However, usually, it's a polygon with a number of end points. You simply write down these points and plug them into your equation Z=x+3y, and whichever one is the minimum is your answer.

Answer 2

The answer above is almost correct. One of the 'right's should have been 'left', or better still 'above'.
By the way, it can be easier not to turn everything into y = mx + b.
I suggest that for each inequality (other than x >0, y>0 and y>ax which are very easy to do anyway) first change it to an equation. Then say if x = 0 what is y and for y = 0 what is x. This gives you two points through which to put a line. You decide which side to shade by considering any point not on the line and whether its values agree with the inequality. Also it is better to shade out the region NOT allowed by the inequality. This leaves clear space for the allowed answer rather than overlapping shading.