Minimize
Z = 5x + 10y
Subject to constraints
x + 2y ≤ 120
x + y ≥ 60
x – 2y ≥ 0
and x , y ≥ 0
Firstly, we draw the graph of the line x + 2y = 120
x | 0 | 120 |
y | 60 | 0 |
Put (0,0) in the inequality x + 2y ≤ 120, we get
0 + 2 × 0 ≤ 120
⇒ 0 ≤ 120 (which is true)
So, half plane is towards the origin.
Secondly, graph of the line x + y = 60
x | 0 | 60 |
y | 60 | 0 |
Put (0,0) in the inequality x + y ≥ 60, we get
0 + 0 ≥ 60
⇒ 0 ≥ 60 (which is false)
So, half plane is away from the origin.
Thirdly, draw the graph of line x – 2y = 0
x | 0 | 10 |
y | 0 | 5 |

On solving equations x – 2y = 0 and x + y = 60, we get B(40, 20)
and on solving equations x – 2y = 0 and x + 2y = 120, we get C(60, 30).
Feasible Region is ABCDA.

So, the minimum value of Z is 300 at the point (60, 0)
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