By graphical method, the solution of linear programming problem

\[\text{ Subject } to \text{ 3 } x_1 + 2 x_2 \leq 18\]

\[ x_1 \leq 4\]

\[ x_2 \leq 6\]

\[ x_1 \geq 0, x_2 \geq 0, \text{ is } \]

#### Options

*x*_{1}= 2,*x*_{2}= 0,*Z*= 6*x*_{1}= 2,*x*_{2}= 6,*Z*= 36*x*_{1}= 4,*x*_{2}= 3,*Z*= 27*x*_{1}= 4,*x*_{2}= 6,*Z*= 42

#### Solution

*x*_{1} = 2, *x*_{2} = 6, Z = 36

We need to maximize the function Z = 3*x*_{1}* *+ 5*x*_{2}

First, we will convert the given inequations into equations, we obtain the following equations: 3*x*_{1}* *+ 2*x*_{2} = 18, *x*_{1} = 4, *x*_{2} = 6, *x*_{1} = 0 and* **x*_{2} = 0

Region represented by 3*x*_{1}* *+ 2*x*_{2} ≤ 18:

The line 3*x*_{1}* *+ 2*x*_{2}_{ }= 18 meets the coordinate axes at *A*(6, 0) and *B*(0, 9) respectively. By joining these points we obtain the line 3*x*_{1}* *+ 2*x*_{2}_{ }= 18.

Clearly (0,0) satisfies the inequation 3*x*_{1}* *+ 2*x*_{2}_{ }= 18 .So,the region in the plane which contain the origin represents the solution set of the inequation* *3*x*_{1}* *+ 2*x*_{2}_{ }= 18.

Region represented by *x*_{1} ≤ 4:

The line* **x*_{1} = 4 is the line that passes through *C*(4, 0) and is parallel to the Y axis. The region to the left of the line *x*_{1} = 4 will satisfy the inequation *x*_{1} ≤ 4.

Region represented by *x*_{2} ≤ 6:

The line* **x*_{2} = 6 is the line that passes through *D*(0, 6) and is parallel to the X axis. The region below the line *x*_{2}_{ }= 6 will satisfy the inequation *x*_{2} ≤ 6.

Region represented by *x*_{1} ≥ 0 and* **x*_{2} ≥ 0:

Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations *x*_{1} ≥ 0 and *x*_{2}* *≥ 0.

The feasible region determined by the system of constraints, 3*x*_{1}* *+ 2*x*_{2} ≤ 18,* **x*_{1} ≤ 4*, **x*_{2} ≤ 6,* **x*_{1} ≥ 0, and *x*_{2} ≥ 0, are as followsCorner points are *O*(0, 0), *D*(0, 6), *F*(2, 6), *E*(4, 3) and *C*(4, 0).

The values of the objective function at these points are given in the following table

Points | Value of Z |

O(0, 0) |
3(0)+5(0) = 0 |

D(0, 6) |
3(0)+5(6) = 30 |

F(2, 6) |
3(2)+5(6) = 36 |

E(4, 3) |
3(4)+5(3) = 27 |

C(4, 0) |
3(4)+5(0) = 12 |

We see that the maximum value of the objective function Z is 36 which is at* F*(2, 6).