Solve the LP problem graphically using level curves. (The numbers after the X should be
subscripted) MAX: 2X1 + 5X2 Subject to: 6X1 + 5X2 LESS than or equal to 60 2X1 +3X2
LESS THAN OR EQUAL TO 24 3X1 + 6X2 LESS THAN OR EQUAL TO 48 X1, X2
GREATER THAN OR EQUAL TO ZERO
Maximize 2𝑋1 + 5𝑋2
Subject to
6𝑋1 + 5𝑋2 ≤ 60
2𝑋1 + 3𝑋2 ≤ 24
3𝑋1 + 6𝑋2 ≤ 48
𝑋1 ≥ 0, 𝑋2 ≥ 0
Step 1
Draw the lines 6𝑋1 + 5𝑋2 = 60, 2𝑋1 + 3𝑋2 = 24, and 3𝑋1 + 6𝑋2 = 48.
To draw a line we need two points on the line.
6𝑋1 + 5𝑋2 = 60 passes through the points (0, 12) and (10,0)
2𝑋1 + 3𝑋2 = 24 passes through the points ((0,8) and (12,0)
3𝑋1 + 6𝑋2 = 48 passes through the points (0,8) and (16,0)
The feasible region of the LP is the shaded region.
The corners of the feasible region are O, A, B, C.
2 X1 + 5X2 = 40
To find the maximum value of the objective function 𝑧 = 2𝑋1 + 5𝑋2, we can daw the line
𝑧 = 2𝑋1 + 5𝑋2 for different values of z and determine that line for which z is maximum and
at same time at least one point common with the shaded region.
The line 2𝑋1 + 5𝑋2 = 40 passes through the point C(0,8). If we take z > 40, then the line
2𝑋1 + 5𝑋2 = 𝑧 will not intersect the shaded region.
So the Optimal solution is the following
𝑋1 = 0, 𝑋2 = 8, 𝑧 = 40
Alternatively we know that the solution of the LP will be one the corners of the feasible
region. So find out the values of z at the corners.
Corner
O
A
B
C
X1
0
10
7.5
0
X2
0
0
3
8
From the above table, z is maximum (=40) when X1 = 0 and X2 = 8
Z=2x1+5x2
0
20
30
40