Maximise and Minimise Z = 3x – 4y

subject to x – 2y ≤ 0

– 3x + y ≤ 4

x – y ≤ 6

x, y ≥ 0

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.