📌 Example 1. Find the maximum value of *z*, given:

{ | 2x+5y≤253 x+2y≤21x≥0y≥0 |

✍ Solution:

The system of linear inequalities determines a set of

**feasible solutions**. The graph of this set is the **feasible region**.

Graph the feasible region determined by the system

of constraints.

If a linear programming problem has a solution, then the solution is at a vertex of the feasible region.

Maximize the value of *z*= 2*x*+3*y* over the feasible region.

Test the value of

*z*at each of the vertices. The maximum value of

*z*is 19. This occurs at the point (5, 3) or when

*x*=5 and

*y*=3.

📌 Example 2.

Solve the following LPP graphically:

Maximise *Z*=2*x*+3*y*,

subject to *x*+*y*≤4, *x*≥0, *y*≥0

✍ Solution:

The shaded region (OAB) in the [Fig. Ex2] is the feasible region determined by the system of constraints *x*≥0, *y*≥0 and *x*+*y*≤4.

The feasible region OAB is bounded, so, maximum value will occur at a corner point of the feasible region.

Corner Points are O(0, 0), A (4, 0) and B (O, 4).

Evaluate *Z* at each of these comer point.

Corner Point | Value of Z |
---|---|

O(0, 0) | 2(0)+3(0)=0 |

A(4, 0) | 2(4)+3(0)=8 |

B(0, 4) | 2(0)+3(4)=12←maximum |

Hence, the maximum value of

*Z*is 12 at the point (0, 4)

📌 Example 3:

Maximize *Z*=3*x*+4y subject to the constraints: *x*+*y*≤4, *x*≥0, *y*≥0.

✍ Solution:

The feasible region determined by the constraints, *x*+*y*≤4, *x*≥0, *y*≥0, is as follows.

The corner points of the feasible region are O(0, 0), A(4, 0), and B(0, 4). The values of

*Z*at these points are as follows.

Corner point | Z=3x+4y |

O(0, 0) | 0 |

A(4, 0) | 12 |

B(0, 4) | 16→Maximum |

Therefore, the maximum value of *Z* is 16 at the point B(0, 4).

📌 Example 4. Determine the maximum value of *Z*=4*x*+3*y* if the feasible region for an LPP is shown in [Fig. Ex4].

✍ Solution:

The feasible region is bounded. Therefore, maximum of *Z* must occur at the corner point of the feasible region [Fig. Ex4].

Corner Point | Value of Z |
---|---|

O(0, 0) | 4(0)+3(0)=0 |

A(25, 0) | 4(25)+3(0)=100 |

B(16, 16) | 4(16)+3(16)=112←maximum |

C(0, 24) | 4(0)+3(24)=72 |

Hence, the maximum value of

*Z*is 112.

📌 Example 5.

Solve the following linear programming problem graphically:

Maximise *Z*=4*x*+*y* … (1)

subject to the constraints:

*x*+

*y*≤50 … (2)

3

*x*+

*y*≤90 … (3)

*x*≥0,

*y*≥0 … (4)

✍ Solution:

The shaded region in [Fig. Ex5] is the feasible region determined by the system of constraints (2) to (4). We observe that the feasible region OABC is

**bounded**. So, we now use Corner Point Method to determine the maximum value of Z.

The coordinates of the corner points O, A, B and C are (0, O), (30, 0), (20, 30) and (0, 50) respectively. Now we evaluate *Z* at each corner point.

Corner Point | Corresponding value of Z |
---|---|

O (0, 0) C (30, 0) B (20, 30) A (0, 50) |
0 120 ← Maximum 110 50 |

Hence, maximum value of

*Z*is 120 at the point (30, 0).

📌 Example 6:

Maximize *Z*=5*x*+3*y*

subject to 3*x*+5*y*≤15, 5*x*+2*y*≤10, *x*≥0, *y*≥0.

✍ Solution:

The feasible region determined by the system of constraints, 3*x*+5*y*≤15, 5*x*+2*y*≤10, *x*≥0, *y*≥0, is as follows.

The corner points of the feasible region are O(0, 0), A(2, 0), B(0, 3), and C(20/19, 45/19). The values of

*Z*at these corner points are as follows.

Corner point | Z=5x+3y |

O(0, 0) | 0 |

A(2, 0) | 10 |

B(0, 3) | 9 |

C(20/19, 45/19) | (235/19)≅12 [Maximum] |

Therefore, the maximum value of *Z* is 235/19 at the point (20/19, 45/19).

📌 Example 7:

Maximize *Z*=3*x*+2*y*

subject to *x*+2*y*≤10, 3*x*+*y*≤15, x, *y*≥0

✍ Solution:

The feasible region determined by the constraints, *x*+2*y*≤10, 3*x*+*y*≤15, *x*≥0 and *y*≥0, is as follows.

The corner points of the feasible region are A(5, 0), B(4, 3), and C(0, 5). The values of

*Z*at these corner points are as follows.

Corner point | Z=3x+2y |

A(5, 0) | 15 |

B(4, 3) | 18→Maximum |

C(0,5) | 10 |

Therefore, the maximum value of *Z* is 18 at the point (4, 3).

📌 Example 8: **Solving a Linear Programming Problem**

Find the maximum value of the objective function

z=4*x*+6y [Objective function]

where *x*≥0 and *y*≥0, subject to the constraints

–x+y≤11x+y≤272 x+5y≤90 |
}Constraints |

✍ Solution:

The region bounded by the constraints is shown in [Fig. Ex8].

By testing the objective function at each vertex, you obtain

At (0, 0): z=4(0)+6(0)=0At (0, 11): z=4(0)+6(11)=66At (5, 16): z=4(5)+6(16)=116 (Maximum value of z) At (15, 12): z=4(15)+6(12)=132At (27, 0): z=4(27)+6(0)=108. |

So, the maximum value of *z* is 132, and this occurs when *x*=15 and *y*=12.

💪 Graphical Method to Solve Linear Programming Problems 💎