Maximise Z = x + y subject to x + 4y ≤ 8, 2x + 3y ≤ 12, 3x + y ≤ 9, x ≥ 0, y ≥ 0.

Given-


Z = x + y


It is subject to constraints


x + 4y ≤ 8


2x + 3y ≤ 12


3x + y ≤ 9


x ≥ 0, y ≥ 0


We need to maximize Z, subject to the above constraints.


Now let us convert the given inequalities into equation.


We obtain the following equation


x + 4y ≤ 8


x + 4y = 8


2x + 3y ≤ 12


2x + 3y = 12


3x + y ≤ 9


3x + y = 9


x ≥ 0


x=0


y ≥ 0


y=0


The region represented by x + 4y ≤ 8:


The line x + 4y = 8 meets the coordinate axes (8,0) and (0,2) respectively. We will join these points to obtain the line x + 4y = 8. It is clear that (0,0) satisfies the inequation x + 4y ≤ 8. So, the region containing the origin represents the solution set of the inequation x + y ≤ 8.


The region represented by 2x + 3y ≤ 12:


The line 2x + 3y = 12 meets the coordinate axes (6,0) and (0,4) respectively. We will join these points to obtain the line 2x + 3y = 12. It is clear that (0,0) satisfies the inequation 2x + 3y ≤ 12. So, the region containing the origin represents the solution set of the inequation 2x + 3y ≤ 12


The region represented by 3x + y ≤ 9:


The line 3x + y = 9 meets the coordinate axes (3,0) and (0,9) respectively. We will join these points to obtain the line 3x + y = 9. It is clear that (0,0) satisfies the inequation 3x + y ≤ 9. So, the region containing the origin represents the solution set of the inequation 3x + y ≤ 9


Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations.


Plotting these equations graphically, we get



Feasible region is ABCD


Value of Z at corner points A, B, C and D –



So, value of Z is maximum at B (2.54, 1.36), the maximum value is 3.90.


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