The optimal solution is (4,2).
How to minimize an objective function graphically.
Given
Objective function Z = 3x + 2y and constraints:
5x + y => 10
x + y => 6
x + 4y => 12
x => 0, y => 0
The feasible region is the area where all these constraints overlap.
Let's plot the following
Plot the line 5x + y = 10 and shade the region above this line.
Plot the line x + y = 6 and shade the region above this line.
Plot the line x + 4y = 12 and shade the region above this line.
Consider x =>0 and y =>0
The feasible region is where all shaded regions overlap.
From the graph, the possible solution points are (0,9), (4,2) , (12,0)
Let's find the value of the objective function at each point
For (0,9)
Z = 3x + 2y
= 3(0) + 2(9)
Z = 18
For (4,2)
Z = 3(4) + 2(2)
= 12 + 4 = 16
For (12, 0)
Z = 3(12) + 2(0)
= 36 + 0 = 36
The point (4,2) minimizes Z
The optimal solution is (4,2).