First we draw the lines AB and CD whose equations are 3x + 5y = 10 and 5x + 3y = 15 respectively.
The feasible region is OCPBD which is shaded in the graph.
The vertices of the feasible region are 0(0, 0), C(3, 0), P and B (0, 2).
P is the point of intersection of the lines
3x + 5y = 10 … (1)
and 5x + 3y = 15 … (2)
Multiplying equation (1) by 5 and equation (2) by 3, we get
15x + 25y = 50
15x + 9y = 45
On subtracting, we get
16y = 5
∴ y = 5/16
Substituting y = 5/16 in equation (1), we get
The values of objective function z = 6x + 10y at these vertices are
z(O) = 6(0) + 10(0) = 0 + 0 = 0
z(C) = 6(3) + 10(0) = 18 + 0 = 18
The maximum value of z is 20 at P \(\left(\frac{45}{16},\frac{5}{16}\right)\) and B (0, 2) two consecutive vertices.
∴ z has maximum value 20 at each point of line segment PB where B is (0, 2) and P is \(\left(\frac{45}{16},\frac{5}{16}\right)\).
Hence, there are infinite number of optimum solutions.
