Question

P=20x+30y

subject to:
2x+3y>30
2x+y<26
-6x+5y<50
x,y >0

Answer 1

Two ways:
Graphical...
plot the contraint areas 2x+3y=30
2x+y=26
-6x+5y=50
x,y >0
Now we have a nice theorem that states the max, and min, will occur on the boundry where constraint lines meet. So test the profit function 20x+30y at each corner, and you will find the max.

Second method: Analytical
http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=1e692c6f72587b2cbd3e7be018fd8960&title=Linear%20Programming%20Calculator&theme=blue

Answer 2

looks like linear programming.

let's look at each region.
2x + 3y > 30
the boundary is 2x + 3y = 30
when x = 0, y = 10
when y = 0 , x = 15 , so we have (0,10) and (15,0)
plot these points on the axes and draw a dotted line,
shade in the region <b>above</b> this line

2x+y < 26
repeat my method, shade in the region below the line 2x+y = 26

careful with the last one, I will do that one as well.
-6x+y < 50 -----> y < 6x+50
draw the boundary line y = 6x+50
when x=0, y = 50 , when y = 0 x = -25/3
so your two points are (0,50) , (-25/3,0)
shade in the region below that line

Your boundary lines should look like this
http://www.wolframalpha.com/input/?i=plot+2x+%2B+3y+%3D+30,+2x%2By+%3D+26,+y+%3D+6x%2B50

and here the actual intersection of the shaded regions.
http://www.wolframalpha.com/input/?i=plot+2x+%2B+3y%3E+30,+2x%2By+%3C+26,+y+%3C+6x%2B50

notice that P = 20x + 30y has the same slope as 2x + 3y = 30
so the farthest point to the right of your region will be the solution.
Solve the corresponding equations to find that point, plug into
P = 20x + 30y