Question

Solve the following L.P.P. by graphical method:

Minimize: Z = 6x + 2y subject to x + 2y ≥ 3, x + 4y ≥ 4, 3x + y ≥ 3, x ≥ 0, y ≥ 0.

Answer 1

To draw the feasible region, construct the table as follows:

Equation of line	Intercepts	Constraint Type	Feasible region
x + 2y = 3	x = 3, y = `3/2`	≥	Non-origin side
x + 4y = 4	x = 4, y = 1	≥	Non-origin side
3x + y = 3	x = 1, y = 3	≥	Non-origin side

 

Pis point of intersection of the lines x + 2 y = 3 and x + 4y = 4.

Solving this equation, we will get x = 2, y = `1/2`

∴ P = `(2, 1/2)`

Q is the point of intersection of the lines x + 2 y = 3 and 3x + y = 3.

Solving this equation, we will get `x = 3/5, y = 6/5`

∴ Q = `(3/5, 6/5)`

The common shaded region is feasible region with boundary points C(4, 0), P`(2, 1/2)`

 Q = `(3/5, 6/5)` and F(0,3).

Substitute these points in z = 6x + 2y

Points	Value of z = 6x + 2y
C(4, 0)	ZC = 6(4) + 2(0) = 24
P`(2, 1/2)`	ZP = `6(2) + 2(1/2) = 13`
Q = `(3/5, 6/5)`	ZQ = `6(3/5) + 2(6/5) = 6`
F(0,3)	ZF = 6(0) + 2(3) = 6

From the above tabulation, Z is minimum at two points Q and F.

∴ Z_min = 6 at `"Q"(3/5, 6/5)` and F(0,3).

∴ The L.P.P. has infinitely many solutions.