A company sells two different products, *A* and *B*. The two products are produced in a common production process, which has a total capacity of 500 man-hours. It takes 5 hours to produce a unit of *A* and 3 hours to produce a unit of *B*. The market has been surveyed and company officials feel that the maximum number of unit of *A* that can be sold is 70 and that for *B* is 125. If the profit is Rs 20 per unit for the product *A* and Rs 15 per unit for the product *B*, how many units of each product should be sold to maximize profit?

#### Solution

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

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

It takes 5 hours to produce a unit of *A* and 3 hours to produce a unit of *B*.The two products are produced in a common production process, which has a total capacity of 500 man-hours.

\[5x + 3y \leq 500\]

The maximum number of unit of *A* that can be sold is 70 and that for *B* is 125.

\[x \leq 70\]

\[y \leq 125\]

If the profit is Rs 20 per unit for the product *A* and Rs 15 per unit for the product *B. *Therefore, profit *x *units of product *A* and *y *units of product *B* is Rs 20*x* and Rs 15*y *respectively.

Total profit = Z = \[20x + 15y\]

The mathematical formulation of the given problem is

Max Z = \[20x + 15y\]

subject to

\[5x + 3y \leq 500\]

\[x \leq 70\]

\[y \leq 125\]

First we will convert inequations into equations as follows:

5*x* + 3*y* = 500, *x* = 70, *y* = 125, *x* = 0 and *y* = 0

Region represented by 5*x* + 3*y* ≤ 500:

The line 5*x* + 3*y* = 500 meets the coordinate axes at *A*_{1}(100, 0) and \[B_1 \left( 0, \frac{500}{3} \right)\] respectively. By joining these points we obtain the line 5*x* + 3*y* = 500. Clearly (0,0) satisfies the 5*x* + 3*y* = 500. So, the region which contains the origin represents the solution set of the inequation 5*x* + 3*y* ≤ 500.

Region represented by *x* ≤ 70:

The line *x* = 70 is the line passes through *C*_{1}(70, 0) and is parallel to *Y* axis. The region to the left of the line *x* = 70 will satisfy the inequation* x* ≤ 70.

Region represented by *y* ≤ 125:

The line *y* = 125 is the line passes through *D*_{1}(0, 125) and is parallel to *X* axis. The region below the the line *y* = 125 will satisfy the inequation* y* ≤ 125.

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 5*x* + 3*y* ≤ 500, *x* ≤ 70, *y* ≤ 125, *x* ≥ 0 and *y* ≥ 0 are as follows.

The corner points are *O*(0, 0), *D*_{1 }\[\left( 0, 125 \right)\] ,* E*_{1}(25, 125), *F*_{1}(70, 50) and *C*_{1}(70, 0).The values of Z at the corner points are

Corner points | Z = \[20x + 15y\] |

O |
0 |

D_{1} |
1875 |

E_{1} |
2375 |

F_{1} |
2150 |

C_{1} |
1400 |

*E*

_{1}

*A*and 125 units of

*B*should be manufactured.