Solve the following L.P.P graphically:
Maximize: Z = 10x + 25y
Subject to: x ≤ 3, y ≤ 3, x + y ≤ 5, x ≥ 0, y ≥ 0
Solution
First we draw the lines AB, CD and EF whose equations are x = 3, y = 3 and x + y = 5 respectively.
| Line | Equation | Point on the X-axis |
Point on the Y-axis |
Sign | Region |
| AB | x = 3 | A(3,0) | - | ≤ | origin side of line AB |
| CD | y = 3 | - | D(0,3) | ≤ | origin side of line CD |
| EF | x + y = 5 | E(5,0) | F(0,5) | ≤ | origin side of line EF |
The feasible region is OAPQDO which is shaded in the figure.
The vertices of the feasible region are O (0,0), A (3, 0), P, Q and D (0, 3)
P is the point of intersection of the lines x + y = 5 and x = 3
Substituting x = 3 in x + y = 5, we get,
3+ y=5
y = 2
P≡ (3, 2)
Q is the point of intersection of the lines x + y = 5 and y = 3
Substituting y = 3 in x + y = 5, we get,
x + 3 = 5
x = 2
Q ≡ (2,3)
The values of the objective function z = 10x + 25y at these vertices are
Z(O) =10(0)+ 25(0)= 0
Z(A) =10(3) + 25(0) = 30
Z(P) =10(3) + 25(2) = 30 + 50 = 80
Z(Q) =10(2) + 25(3) = 20 + 75 = 95
Z(D) =10(0) + 25(3) =75
Z has max imumvalue 95, when x = 2 and y = 3.
