Using the corner method to maximize z=2x+5y?

Question

Can someone help me out here. I can't seem to find any step by step instructions on how to do this.

I am not familiar feasible sets at all..so need a pretty basic explanation..not looking for an answer here.

Its constraints are:
3x+2y>=24
x+2y<=20
x<=6
x>=0, y>=0

I should add that I do not know how to graph the constraints. This is all new to me .

Answer 1

This problem is impossible (or perhaps there is a misprint). Let me change it to illustrate how to solve this type of problem. Maximize z = 2x + 5y with constraints 3x + 2y >= 24, x + 2y <= 20, x >= 0, and y >= 0.

To plot the constraints, consider replacing the inequalities by equal signs. Plot the graph of 3x + 2y = 24. (An easy way to sketch a line is to put the equation in intercept form. Divide the equation by 24, get x/8 + y/12 = 1. The x-intercept is 8, the y-intercept is 12.) Sketch the line trrough the points (8,0) and (0,12), This line divides the plane into two portions, the points below the line and those above the line. Choose any point not on the line, for example, take the point (9,0), which is above the line. Put this into the original constraint: we have 3x + 2y = 3(9) + 2(0) = 27, so at this point 3x + 2y > 24, so ALL POINTS ABOVE the line satisfy the constraint.

Do the same with x + 2y <=20. The intercept form is x/20 + y/10 = 1. Choose a point, say (21,0) and test the constraint. We find that at (21,0), x + 2y = 21, which is not < 20, so points above the line do not satisfy the constraint. Now we know that the points below the second line satisfy the constraint.

Do the same with x >= 0 and y > = 0. (The points in the first quadrant satisfy x and y positive or zero.) Finally, we must find where these lines intersect. The first two lines intersect at (2,9). Now we have determined the triangular region with corners at (2,9), (8,0), and (20,0).

Only points inside this triangle need to be tested. But the situation is much easier, because we only have to test the function z = 2x + 5y at the three corners. At (2,9), z =49; at (8,0), z = 21; at (20,0), x = 40. Thus, the max value of z is 49.

Answer 2

Use graph paper to graph those constraints. Then find the interestion of the lines that are greater than 0 in both x and y and x <-6. The answer has to be when x >0 and x <=6