Question

Maximize z = −x + 2y subjected to constraints x + y ≥ 5, x ≥ 3, x + 2y ≥ 6, y ≥ 0 is this LPP solvable? Justify your answer.

Answer 1

To draw the feasible region, construct table as follows:

Inequality	x + y ≥ 5	x ≥ 3	x + 2y ≥ 6
Corresponding equation (of line)	x + y = 5	x = 3	x + 2y = 6
Intersection of line with X-axis	(5, 0)	(3, 0)	(6, 0)
Intersection of line with Y-axis	(0, 5)	−	(0, 3)
Region	Non-origin side	Non-origin side	Non-origin side

Shaded portion XABC is the feasible region, whose vertices are A (6, 0), B and C

B is the point of intersection of the lines x + y = 5 and x + 2y = 6

Solving the above equations, we get

B ≡ (4, 1)

C is the point of intersection of the lines x = 3 and x + y = 5

Putting x = 3 in x + y = 5, we get

3 + y = 5

∴ y = 2

∴ C ≡ (3, 2)

Here the objective function is

Z = −x + 2y

∴ Z at A(6, 0) = −6 + 2(0) = −6

Z at B(4, 1) = −4 + 2(1) = −2

Z at C(3, 2) = −3 + 2(2) = 1

Here, the feasible region is unbounded.

So, the objective function does not have finite maximum value i.e. value of objective function Z increases indefinitely and hence the L.P.P. has unbounded solution.