A small firm manufacturers items *A* and *B*. The total number of items *A* and *B* that it can manufacture in a day is at the most 24. Item *A* takes one hour to make while item *B *takes only half an hour. The maximum time available per day is 16 hours. If the profit on one unit of item *A* be Rs 300 and one unit of item *B* be Rs 160, how many of each type of item be produced to maximize the profit? Solve the problem graphically.

#### Solution

Let the firm manufacture *x *items of *A* and *y *items of *B* per day.

Number of items cannot be negative.

Therefore, *x* ≥0 and *y* ≥0.

It is given that the total number of items manufactured per day is at most 24.

∴ *x* + *y* ≤ 24

Item *A* takes 1 hour to make and item *B* takes 0.5 hour to make.

The maximum number of hours available per day is 16 hours.

∴ *x* + 0.5*y* ≤ 16

If the profit on one unit of item *A* be Rs 300 and one unit of item *B* be Rs 160.Therefore, profit gained on *x *items of *A* and *y *items of B is

Rs 300*x* and Rs 160*y* respectively.

∴ Total profit, Z = 300*x* + 160*y*

Thus, the mathematical formulation of the given problem is:

Maximise Z = 300*x* + 160*y*

subject to the constraints

*x*+ *y* ≤ 24

*x* + 0.5*y* ≤ 16

*x*, *y* ≥0

First we will convert inequations into equations as follows:*x* + *y* = 24, *x* + 0.5*y* = 16, *x* = 0 and *y* = 0

Region represented by *x* + *y* ≤ 24:

The line *x* + *y* = 24 meets the coordinate axes at *A*_{1}(24, 0) and *B*_{1}(0, 24) respectively. By joining these points we obtain the line *x* + *y* = 24. Clearly (0,0) satisfies the *x* + *y* = 24. So, the region which contains the origin represents the solution set of the inequation*x* + *y* ≤ 24.

Region represented by *x** *+ 0.5*y* ≤ 16:

The line *x* + 0.5*y* = 16 meets the coordinate axes at *C*_{1}(16, 0) and *D*_{1}(0, 32) respectively. By joining these points we obtain the line* x* + 0.5*y* = 16. Clearly (0,0) satisfies the inequation *x** *+ 0.5*y* ≤ 16. So,the region which contains the origin represents the solution set of the inequation *x** *+ 0.5*y* ≤ 16.

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* + *y* ≤ 24, *x** *+ 0.5*y* ≤ 16, *x* ≥ 0and *y* ≥ 0 are as follows.

The corner points are *O*(0, 0) ,*C*_{1}(16, 0), *E*_{1}(8, 16) and *B*_{1}(0, 24).

The value of Z at these corner points are as follows:

Corner point | Z = 300x + 160y |

O(0, 0) |
0 |

C_{1}(16, 0) |
4800 |

E_{1}(8, 16) |
4960 |

B_{1}(0, 24) |
3840 |

Thus, the maximum value of* *Z is 4960 at *E*_{1}(8, 16).

Thus, 8 units of item* A* and 16 units of item *B* should be manufactured per day to maximise the profits.