Let us consider the constraint x_{1} + *x _{2} *= 1. The feasible region of this constraint comprises the set of points on the straight line x

_{1}+

*x*= 1.

_{2}

If the constraint is x_{1} + *x _{2} *≥ 1, then the feasible region comprises not only the set of points on the straight line x

_{1}+

*x*= 1 but also the points above the line. Here above means away from origin.

_{2}

If the constraint is x1 + *x2 *≤ 1, then the feasible region comprises not only the set of points on the straight line *x *1 + *x2 *= 1 but also the points below the line. Here below means towards the origin.

The above three cases depicted below :

For the constraints *x1 *≤ 1 , *x1 *≥ 1,* x2 *≤ 1 , *x2 *≥ 1 the feasible regions are depicted below :

For the constraints x 1 - *x2 *= 0, x 1 - *x2 *≥ 0 and x 1 - *x2 *≥ 0 the feasible regions are depicted in Fig. 1.5.

The *steps of graphical method *can be stated as follows :

( i) Plot all the constraints and identify the individual feasible regions.

(ii) Identify the common feasible region and identify the corner points *i.e., *vertices of the common feasible region.

*(iii) *Identify the optimal solution at the corner points if exists.

**Example 1.** *Using graphical method solve the following LPP :*

*Maximize z = 5x _{1} *+

*3x*

_{2}*Subject to, 2x _{1 }*+

*5x*≤

_{2}*10,*

*5x _{1} *+

*2x*≤

_{2 }*10,*

*2x _{1} *+

*3x*

_{2}≥ 6,*x1 ≥ 0, x2 ≥ 0.*

** **

**Solution.** Let us present all the constraints in intercept form *i.e.,*

The common feasible region ABC is shown

in Fig. 1.6 and the individual regions are

indicated by arrows.

(Due to non-negativity constraints *i.e ..*

*x1 ≥* 0, x2 ≥ 0, the common feasible region is

obtained in the first quadrant).

The corner points are A(18/11 , 10 /11),

B ( 10/7 , 10/7) and C (0, 2). The value of

the objective function at the corner points are

z* _{A}=120 / 11= *10.91,

z* _{B}* =80/7 = 11.43 and z

*= 6.*

_{c}Here the common feasible region is bounded and the maximum has occurred at the corner point

B. Hence the optimal solution is

x1* = 10/7 , *x2* *=10/7 and *z* *=80/7 =11.43.

**Example **2. *Using graphical method solve the following LPP :*

*Minimize z *= *3x _{1} *+

*10x*

_{2}*Subject to, 3x _{1 }*+

*2x*≥

_{2 }*6,*

*4x1 *+ x2 ≥ *4,*

*2x1 *+ *3x2 *≥ *6,*

*x1 *≥ *0, x2 *≥ *0.*

**Solution. **Let us present all the constraints in intercept form *i.e.,*

Due to the non-negativity constraints i.e., *x1*≥0 and x2 ≥ 0 the feasible region will be in the first quadrant.

The conm1on feasible region is shown in Fig. 1.7 where the individual feasible regions are shown by arrows. Here the common feasible region is unbounded.

i.e., open with the comer points A(3, 0), B (3/5,8/5), c(2/5, 12/5) and D (0, 4). The value of the o objective function at corner points are *z _{A} *= 9 ,

*z*= 89/5 = 17.8,z

_{B}_{c}= 126/5 =25.2, an

*z*= 40.

_{D}

Here the minimum has occurred at A and there is no other point in the feasible region at which the objective function value is lower than 9. Hence the optimal solution is

*x1* *= 3, *x2* *= 0 an *z *= 9

** **

**Example **3. *Solve the following LPP by graphical method :*

*Maximize z *= *3x _{1} *-

*15x*

_{2}*Subject to, x1 *+ *x2 *≥*8,*

x1 -4x2 ≥8,

*x 1*≥*0. x2 unrestricted in sign.*

** **

**Solution. **Since x2 is unrestricted in sign this means x2 may be ≥ 0 or ≤ 0. Also *x *1 ≥ 0. Then the common feasible region will be in the first and fourth quadrant. Let us present all the constraints in intercept fom1s *i.e.,*

The common feasible region is shown in Fig. 1.8 where the individual feasible regions are shown by arrows

The value of the objective function at the corner points are *z _{A} *= 30, z

_{B}= 24 and

*z*= -120. Since the common feasible region is bounded and the maximum has occurred at A, the optimal solution is

_{c}*x*_{1}** = *0, *x _{2}* *=- 2 an

*z *=*30.

**3(a) Exceptional Cases in Graphical Method**

There are three cases may arise. When the value of the objective function is maximum/minimum at more than one corner points then *'multiple optima' *solutions are obtained.

Sometimes the optimum solution is obtained at infinity, then the solution is called *' unbounded solution' . *Generally, this type of solution is obtained when the common feasible region is unbounded and the type of the objective function leads to unbounded solution.

When there does not exist any common feasible region, then there does not exist any solution. Then the given LPP is called *infeasible i.e. , *having *no solution. *For example, consider the LPP which is infeasible

Maximize z = *5x _{1} *+ 10x

_{2}

Subject to, *x1 *+ x2 ≥ 2,

*x 1 *+ x2 ≥ 3,

*x 1, x2 *≥ 0.