Minimize Z = 3x + 2y Subject to the constraints x + y ≥ 8, 3x + 5y ≤ 15 and x ≥ 0,y ≤ 15

Question

Solve the linear programming problem by graphical method:

Minimize Z = 3x + 2y Subject to the constraints x + y ≥ 8, 3x + 5y ≤ 15 and x ≥ 0,y ≤ 15

Answer 1

Converting the given in equations into the equations :

x + y = 8 …..(1)

3x + 5y = 15 …..(2)

y = 15 …..(3)

Region represented by x + y ≥ 8 : 

The line x + y = 8 meets the coordinate axis at C(8,0) and D(0, 8).

Table for x + y = 8

X	8	0
y	0	8

A(8, 0); B(0, 8)

Join the points C and D to obtain the line. 

We find that the point (0, 0) does not satisfy the in equation x + y > 8. 

So, the region opposite to the origin represents the solution set to the in equation.

Region represented by 3x + 5y ≤ 15 : 

The line 3x + 5y = 15 meets the coordinate axis at C(5,0) and D(0, 3).

Table for 3x + 5y = 15

X	5	0
y	0	3

C(5,0);D(0,3)

Join the points C and D to obtain the line. 

Clearly (0,0) satisfies the in equation 3x + 5y ≤ 15.

So, the region containing the origin represents the solution set of this in equation.

Region represented by y ≤ 15 : 

Line y = 15 is parallel to x-axis, its each point will satisfy the in equation in first quadrant. 

So, its solution region will be towards origin.

Region represented by x ≥ 0 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 ≥ 0 and y ≥ 0.

There is no any common region for solution.

Hence there is no feasible solution of this problem.