linear programming

Question

Solve this linear programming problem by the methods of corners.  Show the feasible region as a shaded region in the x-y plane.

Maximize P = 3x + 2y

Subject to:  x + Y < or equal to 6

                  x < or equal to 3

                  x> or equal to 0  and Y > or equal to 0

 

 

 

Find the solution set for the system y> 3x + 4 and y < x.  Show the solution set as a shaded area of the x-y plane.

Answer 1

To find the optimum value of P, graph the inequalities.  Since the values of x and y are limited to non negative ones, the solution lies in quadrant I.  The intersection of the various inequalities are (0, 0), (0, 6), (3, 3), and (3, 0). After testing these three sets of values, point (3, 3) gives the highest value of P:  P = 3(3) + 2(3) = 15.

 

For the second part of the problem, plot the equalities as dashed lines, then shade above the first line and below the second line.  The area of intersection lies in quadrant III.