Q. 255.0( 1 Vote )

Maximise and Mini

Answer :


We have constraints,


x – 2y ≤ 0


– 3x + y ≤ 4


x – y ≤ 6


x, y ≥ 0


Z = 3x – 4y


We need to maximize and minimize Z, subject to the given constraints.


Now let us convert the given inequalities into equation.


We obtain the following equation


x – 2y ≤ 0


x - 2y = 0


– 3x + y ≤ 4


-3x + y = 4


x – y ≤ 6


x - y = 6


x ≥ 0


x=0


y ≥ 0


y=0


The region represented by x – 2y ≤ 0:


The line x - 2y = 0 meets the coordinate axes at origin and slope of the line is . We will construct a line passing through origin and whose slope is . As point (1,1) satisfies the inequality. So, the side of line which contains (1,1) is feasible. Hence, the solution set of the inequation x – 2y ≤ 0 is the side which contains (1,1).


The region represented by – 3x + y ≤ 4:


The line – 3x + y = 4 meets the coordinate axes and (0,4) respectively. We will join these points to obtain the line x + y = 7. It is clear that (0,0) satisfies the inequation – 3x + y ≤ 4. So, the region containing the origin represents the solution set of the inequation – 3x + y ≤ 4.


The region represented by x – y ≤ 6:


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


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



The feasible region is region between line -3x + y = 4 and x – y = 6, above BC and to the right of y – axis as shown.


Feasible region is unbounded.


Corner points are A, B, C


So, maximum value of Z at corner points is 12 at C and minimum is -16 at A.


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



So, to check if the solution is correct, we plot 3x – 4y > 12 and 3x – 4y < -16 for maximum and minimum respectively.


The region represented by 3x – 4y > 12:


The line 3x – 4y = 12 meets the coordinate axes (4,0) and (0,-3) respectively. We will join these points to obtain the line 3x – 4y > 12. It is clear that (0,0) does not satisfy the inequation 3x – 4y > 12. So, the region not containing the origin represents the solution set of the inequation 3x – 4y > 12.


The region represented by 3x – 4y <-16:


The line 3x – 4y = -16 meets the coordinate axes and (0,4) respectively. We will join these points to obtain the line 3x – 4y <-16. It is clear that (0,0) does not satisfy the inequation 3x – 4y <-16. So, the region not containing the origin represents the solution set of the inequation 3x – 4y <-16.


We get,



Clearly, 3x – 4y = 12 has no point inside feasible region, but 3x -4y = -16 passes through the feasible region.


Therefore, Z has no minimum value it has only a maximum value which is 12.


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