Maximize Z = -x + 2y Subject to the constraints x ≥ 3, x + y ≥ 5, x + 2y ≥ 6 and x ≥ 0,y ≥ 0

Question

Solve the linear programming problem by graphical method :

Maximize Z = -x + 2y Subject to the constraints x ≥ 3, x + y ≥ 5, x + 2y ≥ 6 and x ≥ 0,y ≥ 0

Answer 1

Converting the given in equations into equations 

x = 3 …..(1)

 x +y = 5 …..(2) 

x + 2y = 6 …..(3) 

y = 0 …..(4) 

Region represented by x + y ≥ 5 : 

The line x + y = 5 meets the coordinate axis on points A(5, 0) and B(0, 5). x + y = 5

x	5	0
y	0	5

A(5,0);B(0,5) Join the points A to 5 to obtain the line.

Clearly (0, 0) does not satisfy the in equation 0 + 0 = 0 ≥ 5. 

So the opposite region to the origin represents the feasible solution region. 

Region represented by x + 2y ≥ 6 : 

The line x + 2y = 6 meets the coordinate axis at point C(6,0) and D(0, 3). 

x + 2y = 6

x	6	0
y	0	3

C(6,0);D(0,3) Join point C to D to obtain the line. 

Clearly (0,0) does not satisfy the in equation 0 + 2(0) = 0 ≥ 6. 

So the region opposite to the origin represents the solution region of the in equation. 

Region represented by x ≥ 3 and y ≥ 0 : 

Since every point in the first quadrant satisfies these in equations. 

So the first quadrant is the region represented by the in equations x ≥ 3 and y ≥ 0.

It is clear from the graph that there is no any common region of the given in equations. 

Hence there does not exist any maximum value for the given in equations.