A carpenter makes chairs and tables profits are ₹ 140 per chair and ₹ 210 per table Both products are processed on three machines, Assembling, Finishing and Polishing the time required for each product in hours and availability of each machine is given by following table.

Product/Machines | Chair (x) |
Table (y) |
Available time (hours) |

Assembling | 3 | 3 | 36 |

Finishing | 5 | 2 | 50 |

Polishing | 2 | 6 | 60 |

Formulate and solve the following Linear programming problems using graphical method.

#### Solution

Let x be the number of chairs and y be the number of tables.

∴ The constraints are

3x + 3y ≤ 36

5x + 2y ≤ 50

2x + 6y ≤ 60

Since x and y are the number of chairs and tables respectively.

∴ They cannot be negative.

∴ x ≥ 0, y ≥ 0

Now, profit for one chair is ₹ 140 and profit for one table is ₹ 210

∴ Total profit (Z) = 140x + 210y

This is objective function to be maximized

∴ Given problem can be formulated as

Maximize Z = 140x + 210y

Subject to 3x + 3y ≤ 36

5x + 2y ≤ 50

2x + 6y ≤ 60

x ≥ 0, y ≥ 0

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

Inequation | Equation | Double intercept form | Points (x, y) | Region |

3x + 3y ≤ 36 | 3x + 3y = 36 | `x/(2) + y/(12)` = 1 | A (12, 0) B (0, 12) |
3(0) + 3(0) ≤ 36 ∴ 0 ≤ 36 ∴ Origin-side |

5x + 2y ≤ 60 | 2x + 6y = 60 | `x/(10) + y/(25)` = 1 | C 10, 0) D 0, 25) |
5(0) + 2(0) ≤ 50 ∴ 0 ≤ 50 ∴ Origin-side |

2x + 6y ≤60 | 2x + 6y = 60 | `x/(30) + y/(10)` = 1 | E (30, 0) F (0, 10 |
2(0) + (0) ≤ 60 ∴ 0 ≤ 60 ∴ Origin-side |

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

y ≥ 0 | y = 0 | – | – | above X-axis |

Shaded portion OFG HC is the feasible region,

Whose vertices are O (0, 0), F (0, 10), G, H and C (10, 0)

G is point of intersection of lines.

2x + 6y = 60

i.e., x + 3y = 30 …(i)

and 3x + 3y = 36

i.e., x + y = 12 …(ii)

∴ By (i) – (ii), we get

x + 3y = 30

x + y = 12

– – –

2y = 18

∴ y = 9

Substituting y = 9 in (ii), we get

x + 9 = 12

∴ x = 12 – 9

∴ x = 3

∴ G = (3, 9)

H is the point of intersection of lines.

3x + 3y = 36

i.e., x + y = 12 …(ii)

5x + 2y = 50 …(iii)

∴ By 2 x (ii) – (iii), we get

2x + 2y = 24

5x + 2y = 50

– – –

– 3x – 26

∴ x = `(26)/(3)`

Substituting x = `(26)/(3)` in (ii), we get

`(26)/(3) + y` = 12

∴ y = `12 - (26)/(3) = (36 - 26)/(3)`

∴ y = `(10)/(3)`

∴ H`(36/3, 10/3)`

Here, the objective function is Z = 140x + 210y

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

Feasible Points | The value of Z = 140x + 210y |

O (0, 0) | Z = 140(0) + 210(0) = 0 |

F (0, 10) | Z = 140(0) + 210(10) = 2100 |

G (3, 9) | Z = 140(3) + 210(9) = 420 + 1890 = 2310 |

H`(36/3, 10/3)` | Z = `140(26/3) + 210(10/3) = (3640)/(3) + (2100)/(3)` = 2310 |

C (10, 0) | Z = 140(10) + 210(0) = 1400 |

∴ Z has maximum value 2310 at G (3, 9)

∴ Maximum profit is ₹ 2310, when x = number of chairs = 3, y = number of tables = 9.