**Solve the following problem :**

Maximize Z = 6x + 10y Subject to 3x + 5y ≤ 10, 5x + 3y ≤ 15, x ≥ 0, y≥0

#### Solution

To find the graphical solution, construct the table as follows:

Inequation | equation | Double intercept form | Points (x1, x2) | Points (x1, x2) |

3x + 5y ≤ 10 | 3x + 5y = 10 | `x/(10/3) + y/(2)` = 1 | A `(10/3, 0)` B (0, 2) |
3(0) + 5(0) ≤ 10 |

5x + 3y ≤ 15 | 5x + 3y = 15 | `x/(3) + y/(5)` = 1 | C (3, 0) D (0, 5) |
5(0) + 3(0) ≤ 15 ∴ 0 ≤ 15 ∴ Origin-side |

x ≥ 0 | x = 0 | – | R.H.S. of Y-axis | |

y ≥ 0 | y = 0 | above X-axis |

The shaded portion OCEB is the feasible region.

Whose vertices are O(0, 0), C(3, 0), E and B(0, 2)

E is the point of intersection of lines

3x + 5y = 10 …(i)

5x + 3y = 15 …(ii)

∴ By 5(i) – 3(ii), we get

15x + 25y = 50

15x + 9y = 45

– – –

16y = 5

∴ y = `(5)/(16)`

Substituting y = `(5)/(16)` in (i), we get

`3x+ 5 xx (5)/(16)` = 10

∴ x = `(45)/(16)`

∴ E`(45/16, 5/16)`

Here, the objective function is Z = 6x + 10y

Now, we will find minimum value of Z as follows:

Feasible points | The value of Z = 6x + 10y |

O (0, 0) | Z = 6(0) + 10(0) = 0 |

C (3, 0) | Z = 6(3) + 10(0) = 18 |

E`(5/16, 45/16)` | Z = `6(45/16) + 10(5/16)` = 20 |

B (0, 2) | Z = 6(0) + 10(2) = 20 |

∴ Z has maximum value 20 at all points on the line 3x + 5y = 10 between B (0, 2) and E`(45/16, 5/16)`

∴ There are infinite number of optimum solutions of the given LPP.