Question

Maximize Z = 3x + 3y, if possible,
Subject to the constraints
$$\begin{gathered} x-y\le 1 \\ x+y\ge 3 \\ x,y\ge 0 \end{gathered}$$

Answer 1

First, we will convert the given inequations into equations, we obtain the following equations:
x − y = 1, x + y = 3, x = 0 and y = 0

Region represented by x − y ≤ 1:
The line x − y = 1 meets the coordinate axes at A(1, 0) and B(0, −1) respectively. By joining these points we obtain the line x − y = 1.
Clearly (0,0) satisfies the inequation x + y ≤ 8. So,the region in xy plane which contain the origin represents the solution set of the inequation x − y ≤ 1.

Region represented by x + y ≥ 3:
The line x + y = 3 meets the coordinate axes at C(3, 0) and D(0, 3) respectively. By joining these points we obtain the line x + y = 3.
Clearly (0,0) satisfies the inequation x + y ≥ 3. So,the region in xy plane which does not contain the origin represents the solution set of the inequation x + y ≥ 3.


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

The feasible region determined by the system of constraints x − y ≤ 1, x + y ≥ 3, x ≥ 0 and y ≥ 0 are as follows.



The feasible region is unbounded.We would obtain the maximum value at infinity.
Therefore,maximum value will be infinity i.e. the solution is unboundedm