A manufacturer of patent medicines is preparing a production plan on medicines, *A* and *B*. There are sufficient raw materials available to make 20000 bottles of *A* and 40000 bottles of *B*, but there are only 45000 bottles into which either of the medicines can be put. Further, it takes 3 hours to prepare enough material to fill 1000 bottles of *A*, it takes 1 hour to prepare enough material to fill 1000 bottles of *B* and there are 66 hours available for this operation. The profit is Rs 8 per bottle for *A* and Rs 7 per bottle for *B*. How should the manufacturer schedule his production in order to maximize his profit?

#### Solution

Let *x* bottles of medicine *A* and *y* bottles of medicine *B *are prepared.

Number of bottles cannot be negative.

Therefore,

\[x, y \geq 0\] According to question, the constraints are

\[x \leq 20000\]

\[y \leq 40000\]

\[x + y \leq 45000\]

*A*, it takes 1 hour to prepare enough material to fill 1000 bottles of

*B*

Time taken to fill one bottle of A is \[\frac{3}{1000}\] hrs and time taken by to fill one bottle of B is \[\frac{1}{1000}\] hrs . Therefore, time taken to fill

*x*bottles of A and

*y*bottles of B is \[\frac{3x}{1000}\] hrs and \[\frac{y}{1000}\]hrs respectively.

*A*and Rs 7 per bottle for

*B*. Therefore, profit gained on

*x*bottles of medicine

*A*and

*y*bottles of medicine

*B*is 8

*x*

*and*

*7*

*y*

*respectively.*

Total profit = Z = \[8x + 7y\] which is to be maximised.

Max Z = \[8x + 7y\]

\[x \leq 20000\]

\[y \leq 40000\]

\[x + y \leq 45000\]

*x*=20000,

*y*= 40000,

*x*+

*y*= 45000, 3

*x*+

*y*= 66000,

*x*= 0 and

*y*= 0

Region represented by

*x*≤ 20000:

The line

*x*= 20000 is the line that passes through

*A*

_{1}(20000, 0) and is parallel to

*Y*axis.The region to the left of the

*line*

*x*= 20000 will satisfy the inequation

*x*≤ 20000.

Region represented by

*y*≤ 40000:

The line

*y*= 40000 is the line that passes through

*B*

_{1}(0, 40000) and is parallel to

*X*axis.The region below the

*line*

*y*= 40000 will satisfy the inequation

*y*≤ 40000.

Region represented by

*x*

*+*

*y*≤ 45000:

The line

*x*+

*y*= 45000 meets the coordinate axes at

*C*

_{1}(45000, 0) and

*D*

_{1}(0, 45000) respectively. By joining these points we obtain the line

*x*+

*y*= 45000. Clearly (0,0) satisfies the inequation

*x*

*+*

*y*≤ 45000. So,the region which contains the origin represents the solution set of the inequation

*x*

*+*

*y*≤ 45000.

Region represented by 3

*x*

*+*

*y*≤ 66000:

The line 3

*x*+

*y*= 66000 meets the coordinate axes at

*E*

_{1}(22000, 0) and \[F_1 \left( 0, 66000 \right)\] respectively. By joining these points we obtain the line 3

*x*+

*y*= 66000. Clearly (0,0) satisfies the inequation 3

*x*

*+*

*y*≤ 66000. So,the region which contains the origin represents the solution set of the inequation 3

*x*

*+*

*y*≤ 66000.

Region represented by

*x*≥ 0 and

*y*≥ 0:

Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations

*x*≥ 0, and

*y*≥ 0.

The feasible region determined by the system of constraints

*x*≤ 20000,

*y*≤ 40000,

*x*

*+*

*y*≤ 45000, 3

*x*

*+*

*y*≤ 66000,

*x*≥ 0 and

*y*≥ 0 are as follows.

The corner points are

*O*(0, 0),

*B*

_{1}(0, 40000),

*G*

_{1}(10500, 34500),

*H*

_{1}(6000, 20000) and

*A*

_{1}(20000, 0).

The values of Z at these corner points are as follows

Corner point | Z= 8x + 7y |

O | 0 |

B_{1} |
280000 |

G_{1} |
325500 |

H_{1} |
188000 |

A_{1} |
160000 |

The maximum value of Z is 325500 which is attained at

*G*

_{1}(10500, 34500).

Thus, the maximum profit is Rs 325500 obtained when 10500 bottles of

*A*and 34500 bottles of

*B*were manufactured.