A company produces two types of goods, *A* and *B*, that require gold and silver. Each unit of type *A* requires 3 gm of silver and 1 gm of gold while that of type *B* requires 1 gm of silver and 2 gm of gold. The company can produce 9 gm of silver and 8 gm of gold. If each unit of type *A* brings a profit of Rs 40 and that of type *B* Rs 50, find the number of units of each type that the company should produce to maximize the profit. What is the maximum profit?

#### Solution

Let *x *goods of type *A* and *y* goods of type *B* were produced.

Number of goods cannot be negative.

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

The given information can be tabulated as follows:

Silver(gm) | Gold white(gm) | |

Type A | 3 | 1 |

Type B | 1 | 2 |

Availability | 9 | 8 |

Therefore, the constraints are

\[3x + y \leq 9\]

\[x + 2y \leq 8\]

If each unit of type *A* brings a profit of Rs 40 and that of type *B* Rs 50.Then, *x* goods of type *A* and *y* goods of type *B* brings a profit of Rs 40*x* and Rs 50*y**.*

Total profit = Z = \[40x + 50y\] which is to be maximised.

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

Max Z = \[40x + 50y\]

subject to

\[3x + y \leq 9\]

\[x + 2y \leq 8\]

First we will convert inequations into equations as follows :

3*x* + *y* = 9, *x* + 2*y* = 8,* x* = 0 and *y* = 0

Region represented by 3*x* + *y* ≤ 9:

The line 3*x* + *y* = 9 meets the coordinate axes at *A*_{1}(3, 0) and *B*_{1}(0, 9) respectively. By joining these points we obtain the line

3*x* + *y* = 9. Clearly (0,0) satisfies the* *3*x* + *y* = 9. So,the region which contains the origin represents the solution set of the inequation 3*x* + *y* ≤ 9.

Region represented by *x* + 2*y* ≤ 8:

The line *x* + 2*y* = 8 meets the coordinate axes at *C*_{1}(8, 0) and *D*_{1}(0, 4) respectively. By joining these points we obtain the line *x* + 2*y* = 8. Clearly (0,0) satisfies the inequation *x* + 2*y* ≤ 8. So,the region which contains the origin represents the solution set of the inequation *x* + 2*y* ≤ 8.

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 3*x* + *y* ≤ 9, *x* + 2*y* ≤ 8, *x* ≥ 0, and *y* ≥ 0 are as follows.

The corner points are *O*(0, 0), *D*_{1}(0, 4),* **E*_{1}(2, 3), *A*_{1}(3, 0)

The values of Z at these corner points are as follows

Corner point | Z = 40x + 50y |

O |
0 |

D_{1} |
200 |

E_{1} |
230 |

A_{1} |
120 |

The maximum value of Z is 230 which is attained at *E*_{1}(2, 3).

Thus, the maximum profit is of Rs 230 obtained when 2 units of type *A* and 3 units of type *B *produced.