Question

Consider the linear programming problem: Maximum z = 50x + 40y 

Subject to constraints: x + 2y < 10; 3x + 4y < 24; x, y < 0 

(i) Draw the feasible region 

(ii) Find the comer points of the feasible region 

(iii) Find the maximum value of z.

Answer 1

(i) 

(ii) In the gure the shaded region ABC is the feasible region. Here the region is bounded. The corner points are A(4, 3), B(0, 6), C(0, 5) 

(iii) Given; Z = 50x + 40y

Corner points	Value of Z
A	Z = 50(4)+ 40(3) = 320
B	Z= 50(0)+ 40(6) = 240
C	Z= 50(0)+ 40(5) = 200

Since minimum value of Z occurs at A, the solution is 

Z = 50(4) + 4(3) = 320