A carpenter has 90, 80 and 50 running feet respectively of teak wood, plywood and rosewood which is used to product A and product B. Each unit of product A requires 2, 1 and 1 running feet and each unit of product B requires 1, 2 and 1 running feet of teak wood, plywood and rosewood respectively. If product A is sold for Rs. 48 per unit and product B is sold for Rs. 40 per unit, how many units of product A and product B should be produced and sold by the carpenter, in order to obtain the maximum gross income? Formulate the above as a Linear Programming Problem and solve it, indicating clearly the feasible region in the graph.

#### Solution

Product | A | B | |

Teak wood | 2 | 1 | ≤90 |

plywood | 1 | 2 | ≤80 |

Rosewood | 1 | 1 | ≤50 |

Selling price per unit Rs. 48/- for product A and for product B Rs. 40/Mathematical formation :

`Z_max = 48x_1 + 40x_2` (objective function)

s.t. 2x_{1} + x_{2} ≤ 90

x_{1} + 2x_{2} ≤ 80:

x_{1} + x_{2} ≤ 50;

x_{1},x_{2} ≥ 0

2x_{1} + x_{2} = 90 ...........(1)

x_{1} + 2x_{2} = 80 ..........(2)

X_{1} + x_{2} =50 ........(3)

from eq (3) and (2)

x_{2} = 30

x_{1} = 20

from eq. (1) & (3)

`x_{1} = 40`

`x_{2} = 10`

0(0,0)

A(0,40) | Z_{max }= 48 × x1 + 40 × x_{2} |
Z_{A} = 1600 |

B(20,30) | = 48 × 20 + 40 × 30 | Z_{B} = 2160 |

C(40,10) | = 48 × 40 + 40 +10 | Z_{c} = 2320 |

D(45,0) | = 48 × 45 + 40(0) | Z_{D} = 2160 |

40, 10), ZC = 2320.

Hence the optimal solution is x1 = 40, x2 = 10 (Product A = 40 units, Product B = 10 units)