A manufacturer has three machines installed in his factory. machines I and II are capable of being operated for at most 12 hours whereas Machine III must operate at least for 5 hours a day. He produces only two items, each requiring the use of three machines. The number of hours required for producing one unit each of the items on the three machines is given in the following table:

Item | Number of hours required by the machine | ||

AB |
I | II | III |

1 2 |
2 1 |
1 5/4 |

He makes a profit of Rs 6.00 on item *A* and Rs 4.00 on item *B*. Assuming that he can sell all that he produces, how many of each item should he produces so as to maximize his profit? Determine his maximum profit. Formulate this LPP mathematically and then solve it.

#### Solution

Let *x *units of item *A* and *y *units of item *B* be manufactured.

Therefore, \[x, y \geq 0\]

As we are given,

Item |
Number of hours required by the machine |
||

AB |
I | II | III |

1 2 |
2 1 |
1 5/4 |

Machines I and II are capable of being operated for at most 12 hours whereas Machine III must operate at least for 5 hours a day.

According to question, the constraints are

\[x + 2y \leq 12\]

\[2x + y \leq 12\]

\[x + \frac{5}{4}y \geq 5\]

He makes a profit of Rs 6.00 on item *A* and Rs 4.00 on item *B*.

Profit made by him in producing *x* items of *A* and *y* items of *B* is 6*x* + 4*y*.

Total profit Z = \[6x + 4y\] which is to be maximised

Thus, the mathematical formulation of the given linear programmimg problem is

Max Z =\[6x + 4y\] subject to

\[x + 2y \leq 12\]

\[2x + y \leq 12\]

\[x + \frac{5}{4}y \geq 5\]

\[x, y \geq 0\]

First we will convert inequations into equations as follows :*x* + 2*y* = 12, 2*x* + *y* = 12,

*x*= 0 and

*y*= 0

Region represented by

*x*+ 2

*y*≤ 12:

The line

*x*+ 2

*y*= 12 meets the coordinate axes at

*A*

_{1}(12, 0) and

*B*

_{1}(0, 6) respectively. By joining these points we obtain the line

*x*+ 2

*y*= 12.Clearly (0,0) satisfies the

*x*+ 2

*y*= 12. So, the region which contains the origin represents the solution set of the inequation

*x*+ 2

*y*≤ 12.

Region represented by 2

*x*+

*y*≤ 12:

The line 2

*x*+

*y*= 12 meets the coordinate axes at

*C*

_{1}(6, 0) and

*D*

_{1}(0, 12) respectively. By joining these points we obtain the line 2

*x*+

*y*= 12. Clearly (0,0) satisfies the inequation 2

*x*+

*y*≤ 12. So,the region which contains the origin represents the solution set of the inequation 2

*x*+

*y*≤ 12.

Region represented by \[x + \frac{5}{4}y = 5\]

*E*

_{1}(5, 0) and

*F*

_{1}(0, 4) respectively. By joining these points we obtain the line

*x*≥ 0 and

*y*≥ 0:

Since, every point in the first quadrant satisfies these inequations. So,the region which does not contains the origin represents the solution set of the inequation \[x + \frac{5}{4}y \geq 5\]

*x*≥ 0, and

*y*≥ 0.

The feasible region determined by the system of constraints

*x*+ 2

*y*≤ 12, 2

*x*+

*y*≤ 12,

*x*≥ 0, and

*y*≥ 0 are as follows.

The corner points are

*B*

_{1}(0, 6),

*G*

_{1}(4, 4),

*C*

_{1}(6, 0),

*E*

_{1}(5, 0) and

*F*

_{1}(0, 4).

The values of Z at these corner points are as follows

Corner point | Z = 6x + 4y |

B_{1} |
24 |

G_{1} |
40 |

C_{1} |
36 |

E_{1} |
30 |

F_{1} |
16 |

The maximum value of Z is 40 which is attained at

*G*

_{1}(4, 4).

Thus, the maximum profit is Rs 40 obtained when 4 units each of item

*A*and

*B*are manufactured.