For Linear programming, how would you find the optimal solution or?

Question

I'm unsure how to maximize
p=x+2y
subject to:

30x + 20y<=600
.1x +.4y<=4
.2x + .3y<=4.5
x>= 0, y>= 0


any help would be appreciated

Answer 1

The feasible region is in a plane, so it's easy to give an answer in this case. The theory of linear optimisation says that the maximum (and the minimum) value of the objective function p will be found on the boundary of the feasible region. More specifically, it will occur at one of the vertices of the polygon formed by the feasible region (and possibly on one of the sides, if the max occurs at two adjacent vertices).

So here is an easy way to find the maximum value of p. It helps if you draw the feasible region, to help keep track of what you are doing. Determine first the points of intersection of the constraints (basically, the vertices of the polygon formed), by solving the corresponding equations. You will then have a set of points (x,y) representing the boundary of the feasible region.

Then, just plug each of these points into the objective function p, and see which one gives the maximum value. If two vertices give the same value, then every point on the side between them will also give that value.

Answer 2

Optimal Solution Linear Programming

Answer 3

Your restriction create a polygon in the first quadrant with up to 5 points. You should plot it out to see what this is. The theory says the optimal solution occurs on this polygon perimeter. You can draw contours of p=x+2y where p is some given value. The highest value of p within the polygon is the maximum point.

Answer 4

Hmmm, interesting problem. Since x and y are not on equal footing in the expression for p (the contribution from y is doubled whereas the contribution from x is not), here is my suggestion. Use substitution, let w=2y. Thus, p=x+w. Now x and w are on equal footing (they both contribute to p equally). This means that maximizing p is a simple matter of maximizing x and w.

You will then need to make this substitution in all of the conditional inequalities (w=2y means that y=w/2). Graph the conditional inequalities with x and w (put w on the y-axis), shading in the appropriate region that satisfies all of the inequalities. Find the point that maximizes x + w (it must be one of the corner points). Whatever w happens to equal, replace w with 2y and solve for y (ie y will be half of whatever the maximized value of w is, since y=w/2).

Hope this helps. Thanks for the interesting question.

Answer 5

In both case you can proceed in the following fashion. Start with hypothesis contrary to the statement. Follow step by step logical argument which leads you to the impossibility. Thus proving your hypothesis.

Answer 6

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RE:
For Linear programming, how would you find the optimal solution or?
I'm unsure how to maximize
p=x+2y
subject to:

30x + 20y<=600
.1x +.4y<=4
.2x + .3y<=4.5
x>= 0, y>= 0


any help would be appreciated

Answer 7

graph:

3x + 2y = 60
x + 4y = 40
2x + 3y = 45

feasible region bounded by polygon with corners at
(0,0), (0,10), (12,7), (18,3), (20,0) which you get by solving pairs of equations.

evaluate p at each corner: p(12,7) = 26, p(18,3) = 24, and so on. max at (12,7).

more general process is the simplex method. see link.