**Solve the following problem :**

Maximize Z = 4x_{1} + 3x_{2} Subject to 3x_{1} + x_{2} ≤ 15, 3x_{1} + 4x_{2} ≤ 24, x_{1} ≥ 0, x_{2} ≥ 0

#### Solution

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

Inequation | Equation | Double intercept form | Points (x_{1}, x_{2}) |
Region |

3x_{1} + x_{2} ≤ 15 |
3x_{1} + x_{2} = 15 |
`x_1/(5) + x_2/(15)` = 1 | A (5, 0) B (0, 15) |
3(0) + 0 ≤ 15 ∴ 0 ≤ 15 ∴ origin side |

3x_{1} + 4x_{2} ≤ 24 |
3x_{1} + 4x_{2} = 24 |
`x_1/8 + x_2/(6)` = 1 | C (8, 0) D (0, 6) |
3(0 + 4(0) ≤ 24 ∴ 0 ≤ 24 ∴ origin side |

x_{1} ≥ 0 |
x_{1} = 0 |
– | – | R.H.S. of Y-axis |

x_{2} ≥ 0 |
x_{2} = 0 |
– | – | above X-axis |

Shaded portion ODEA is the feasible region.

Whose vertices are O (0, 0), D (0, 6), E, A (5, 0)

E is the point of intersection of the lines

3x_{1} + x_{2} = 15 …(i)

and 3x_{1} + 4x_{2} = 24 …(ii)

∴ By (i) – (ii), we get

3x_{1} + x_{2} = 15

3x_{1} + 4x_{2} = 24

– – –

–3x_{2} = – 9

∴ x_{2} = `(-9)/(-3)`

∴ x_{2} = 3

Substituting x_{2} = 3 in i, we get

3x_{1} + 3 = 15

∴ 3x_{1} = 15 – 3

∴ 3x_{1} = 12

∴ x_{1} = `(12)/(3)` = 4

∴ E (4, 3)

Here, the objective function is Z = 4x_{1} + 3x_{2}

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

Feasible points | The value of Z = 4x_{1} + 3x_{2} |

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

D (0, 6) | Z = 4(0) + 3(0) = 0 |

E (4, 3) | Z = 4(4) + 3(3) = 16 + 9 = 25 |

A (5, 0) | Z = 4(4) + 3(3) = 16 + 9 = 25 |

∴ Z has maximum value 25 at E(4, 3)

∴ Z is maximum, when x_{1} = 4, x_{2} = 3.