Question

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?

Answer 1

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

\[x \leq 20000\]
\[y \leq 40000\]
\[x + y \leq 45000\]  

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
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.

It is given that there are 66 hours available for this operation.

\[\therefore \frac{3x}{1000} + \frac{y}{1000} \leq 66\]

The profit is Rs 8 per bottle for A and Rs 7 per bottle for B.   Therefore, profit gained on xbottles of medicine A and y bottles of medicine B  is 8x and 7y respectively.
Total profit = Z = \[8x + 7y\]  which is to be maximised.

Thus, the mathematical formulat​ion of the given linear programmimg problem is 
Max Z =  \[8x + 7y\]

subject to
\[x \leq 20000\]
\[y \leq 40000\]
\[x + y \leq 45000\]

\[\frac{3x}{1000} + \frac{y}{1000} \leq 66 \Rightarrow 3x + y \leq 66000\]

\[x, y \geq 0\]

First we will convert inequations into equations as follows:
x =20000, y = 40000, x + y = 45000, 3x + y = 66000, x = 0 and y = 0

Region represented by x  ≤ 20000:
The line x  = 20000 is the line that passes through A₁(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 B₁(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 C₁(45000, 0) and D₁(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 3x + y ≤ 66000:
The line 3x + y = 66000 meets the coordinate axes at E₁(22000, 0) and \[F_1 \left( 0, 66000 \right)\] respectively. By joining these points we obtain the line 3x + y = 66000. Clearly (0,0) satisfies the inequation 3x + y ≤ 66000. So,the region which contains the origin represents the solution set of the inequation 3x + 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, 3x + y ≤ 66000, x ≥ 0 and y ≥ 0 are as follows.
The corner points are O(0, 0), B₁(0, 40000), G₁(10500, 34500), H₁(6000, 20000) and A₁(20000, 0). 

The values of Z at these corner points are as follows
 

Corner point	Z= 8x + 7y
O	0
B1	280000
G1	325500
H1	188000
A1	160000

The maximum value of Z is 325500 which is attained at G₁(10500, 34500).
Thus, the maximum profit is Rs 325500 obtained when 10500 bottles of A and 34500 bottles of B were manufactured.