#### Question

A company produces two types of leather belts, say type *A* and *B*. Belt *A* is a superior quality and belt *B* is of a lower quality. Profits on each type of belt are Rs 2 and Rs 1.50 per belt, respectively. Each belt of type *A* requires twice as much time as required by a belt of type *B*. If all belts were of type *B*, the company could produce 1000 belts per day. But the supply of leather is sufficient only for 800 belts per day (both *A* and *B* combined). Belt *A* requires a fancy buckle and only 400 fancy buckles are available for this per day. For belt of type *B*, only 700 buckles are available per day.

How should the company manufacture the two types of belts in order to have a maximum overall profit?

#### Solution

Let the company produces *x* belts of type *A* and *y* belts of type *B*.

Number of belts cannot be negative.

Therefore,

*A*and

*B*combined). Therefore,

*x*+

*y*≤ 800

It is given that the rate of production of belts of type B is 1000 per day.Hence, the time taken to produce y belts of type B is \[\frac{y}{1000}\] And, since each belt of type

*A*requires twice as much time as a belt of type

*B*, the rate of production of belts of type

*A*is 500 per day and therefore, total time taken to produce

*x*belts of type

*A*is \[\frac{x}{500}\] Thus, we have \[\frac{x}{500} + \frac{y}{1000} \leq 1\] \[ \Rightarrow 2x + y \leq 1000\] Belt

*A*requires a fancy buckle and only 400 fancy buckles are available for this per day.

*x*≤ 400

For belt of type

*B*, only 700 buckles are available per day.

*y*≤ 700

Profits on each type of belt are Rs 2 and Rs 1.50 per belt, respectively. Therefore, profit gained on

*x*belts of type

*A*and

*y*belts of type

*B*is Rs 2

*x*and

Rs 1.50

*y*respectively.Hence, the total profit would be Rs (2

*x*+ 1.50

*y*).

Let Z denote the total profit.

Thus, the mathematical formulation of the given linear programming problem is Max Z = \[2x + 1 . 5y\] subject to \[x + y \leq 800\]

\[2x + y \leq 1000\]

\[x \leq 400\]

\[y \leq 700\]

\[x, y \geq 0\]

First we will convert inequations into equations as follows :

*x*+

*y*= 800, 2

*x*+

*y*= 1000,

*x*= 400,

*y*= 700

*, x*= 0 and

*y*= 0

Region represented by

*x*+

*y*≤ 800:

The line

*x*+

*y*= 800 meets the coordinate axes at

*A*

_{1}(800, 0) and

*B*

_{1}(0, 800) respectively. By joining these points we obtain the line

*x*+

*y*= 800. Clearly (0,0) satisfies the

*x*+

*y*= 800. So, the region which contains the origin represents the solution set of the inequation

*x*+

*y*≤ 800.

Region represented by 2

*x*+

*y*≤ 1000:

The line 2

*x*+

*y*= 1000 meets the coordinate axes at

*C*

_{1}(500, 0) and

*D*

_{1}(0, 1000) respectively. By joining these points we obtain the line 2

*x*+

*y*= 1000. Clearly (0,0) satisfies the inequation 2

*x*+

*y*≤ 1000. So,the region which contains the origin represents the solution set of the inequation 2

*x*+

*y*≤ 1000.

Region represented by

*x*≤ 400:

The line

*x*= 400 will pass through

*E*

_{1}(400, 0). The region to the left of the line

*x*= 400 will satisfy the inequation

*x*≤ 400.

Region represented by

*y*≤ 700:

The line

*y*= 700 will pass through

*F*

_{1}(0, 700). The region below the line

*y*= 700 will satisfy the inequation

*y*≤ 700.

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*≤ 800, 2

*x*+

*y*≤ 1000,

*x*≤ 400,

*y*≤ 700,

*x*≥ 0, and

*y*≥ 0 are as follows.

The feasible region determined by the system of constraints isThe corner points are

*F*

_{1}(0, 700),

*G*

_{1}(200, 600),

*H*

_{1}(400, 200) and

*E*

_{1}(400, 0).

The values of Z at these corner points are as follows

Corner point | Z= 2x +1.5y |

F_{1}(0, 700) |
1050 |

G_{1}(200, 600) |
1300 |

H_{1}(400, 200) |
1100 |

E_{1}(400, 0) |
800 |

The maximum value of Z is 1300 which is attained at

*G*

_{1}(200, 600).

Thus, the maximum profit is Rs 1300 obtained when 200 belts of type

*A*and 600 belts of type

*B*were produced.