A manufacturer makes two products, *A* and *B*. Product *A* sells at Rs 200 each and takes 1/2 hour to make. Product *B* sells at Rs 300 each and takes 1 hour to make. There is a permanent order for 14 units of product *A* and 16 units of product *B*. A working week consists of 40 hours of production and the weekly turn over must not be less than Rs 10000. If the profit on each of product *A* is Rs 20 and an product *B* is Rs 30, then how many of each should be produced so that the profit is maximum? Also find the maximum profit.

#### Solution

Let *x* units of product *A* and *y* units of product *B* were manufactured.

Number of units cannot be negative.

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

According to question, the given information can be tabulated as:

Selling price(Rs) | Manufacturing time(hrs) | |

Product A(x) |
200 | 0.5 |

Product B(y) |
300 | 1 |

Also, the availability of time is 40 hours and the revenue should be atleast Rs 10000.

Further, it is given that there is a permanent order for 14 units of product A and 16 units of product B.

Therefore, the constraints are

\[200x + 300y \geq 10000\]

\[0 . 5x + y \leq 40\]

\[x \geq 14\]

\[y \geq 16\]

If the profit on each of product *A* is Rs 20 and on product *B* is Rs 30.Therefore, profit gained on *x* units of product *A* and *y* units of product *B *is Rs 20*x** *and Rs 30*y* respectively.

Total profit = Z = \[20x + 30y\] which is to be maximised

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

Max Z = \[20x + 30y\]

subject to

\[2x + 3y \geq 100\]

\[x + 2y \leq 80\]

\[x \geq 14\]

\[y \geq 16\]

First we will convert inequations into equations as follows:

2*x* + 3*y* = 100, *x* + 2*y* = 80, *x* = 14, *y* = 16, *x* = 0 and *y* = 0.

Region represented by 2*x* + 3*y* ≥ 100:

The line 2*x* + 3*y* = 100 meets the coordinate axes at *A*_{1}(50, 0) and \[B_1 \left( 0, \frac{100}{3} \right)\] respectively. By joining these points we obtain the line

2*x* + 3*y* = 100 . Clearly (0,0) does not satisfies the 2*x* + 3*y* = 100. So, the region which does not contain the origin represents the solution set of the inequation 2*x* + 3*y* ≥ 100.

Region represented by *x** *+ 2*y* ≤ 80:

The line *x** *+ 2*y* = 80 meets the coordinate axes at *C*_{1}(80, 0) and *D*_{1}(0, 40) respectively. By joining these points we obtain the line*x** *+ 2*y* = 80. Clearly (0,0) satisfies the inequation *x** *+ 2*y* ≤ 80. So,the region which contains the origin represents the solution set of the inequation *x** *+ 2*y* ≤ 80.

Region represented by *x* ≥ 14*x* = 14 is the line passes through (14, 0) and is parallel to the Y axis.The region to the right of the line *x* = 14 will satisfy the inequation.

Region represented by *y* ≥ 16*y* = 16 is the line passes through (0, 16) and is parallel to the X axis.The region above the line *y* = 16 will satisfy the inequation.

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 2*x* + 3*y* ≥ 100, *x** *+ 2*y* ≤ 80, *x*≥ 14, *y* ≥ 16, *x* ≥ 0 and *y* ≥ 0 are as follows.

The corner points of the feasible region are *E*_{1}(26, 16), *F*_{1}(48, 16), *G*_{1}(14, 33) and *H*_{1}(14, 24)

The values of Z at these corner points are as follows

Corner point | Z= 20x + 30y |

E_{1} |
1000 |

F_{1} |
1440 |

G_{1} |
1270 |

H_{1} |
1000 |

The maximum value of Z is Rs 1440 which is attained at *F*_{1}

*A*and 16 units of product

*B*were manufactured.