A factory uses three different resources for the manufacture of two different products, 20 units of the resources *A*, 12 units of *B* and 16 units of *C* being available. 1 unit of the first product requires 2, 2 and 4 units of the respective resources and 1 unit of the second product requires 4, 2 and 0 units of respective resources. It is known that the first product gives a profit of 2 monetary units per unit and the second 3. Formulate the linear programming problem. How many units of each product should be manufactured for maximizing the profit? Solve it graphically.

#### Solution

Let *x* units of first product and *y* units of second product be manufactured.

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

The given information can be tabulated as follows:

Product | Resource A |
Resource B |
Resource C |

First(x) |
2 | 2 | 4 |

Second(y) |
4 | 2 | 0 |

Availability | 20 | 12 | 16 |

Therefore, the constraints are

\[2x + 4y \leq 20\]

\[2x + 2y \leq 12\]

\[4x + 0y \leq 16 \text{ or } 4x \leq 16\]

It is known that the first product gives a profit of 2 monetary units per unit and the second 3. Therefore, profit gained from *x* units of first product and *y* units of second product is 2*x** *monetary units and 4*y** *monetary units respectively.

Total profit = Z = \[2x + 3y\] which is to be maximised

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

Max Z = \[2x + 3y\]

subject to

\[2x + 4y \leq 20\]

\[2x + 2y \leq 12\]

\[4x + 0y \leq 16 \text { or} 4x \leq 16\]

\[x, y \geq 0\]

First we will convert inequations into equations as follows :

2*x* + 4*y* = 20, 2*x* + 2*y* = 12, 4*x* = 16,* x* = 0 and *y* = 0

Region represented by 2*x* + 4*y* ≤ 20:

The line 2*x* + 4*y* = 20 meets the coordinate axes at *A*_{1}(10, 0) and *B*_{1}(0, 5) respectively. By joining these points we obtain the line 2*x* + 4*y* = 20. Clearly (0,0) satisfies the* *3*x* + 2*y* = 210. So,the region which contains the origin represents the solution set of the inequation 2*x* + 4*y* ≤ 20.

Region represented by 2*x* + 2*y* ≤ 12:

The line 2*x* +2*y* =16 meets the coordinate axes at *C*_{1}(6, 0) and* **D*_{1}(0, 6) respectively. By joining these points we obtain the line 2*x* + 2*y* = 12. Clearly (0,0) satisfies the inequation 2*x* + 2*y* ≤ 12. So, the region which contains the origin represents the solution set of the inequation 2*x* + 2*y* ≤ 12.

Region represented by 4*x* ≤ 16:

The line 4*x* =16 or *x* = 4 is the line passing through the point *E*_{1}(4, 0) and is parallel to *Y *axis.The region to the left of the line x = 4 would satisfy the inequation 4*x* ≤ 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 2*x* + 4*y* ≤ 20, 2*x* + 2*y* ≤ 12, 4*x* ≤ 16, *x* ≥ 0 and *y* ≥ 0 are as followsThe corner points are *O*(0, 0), *B*_{1}(0, 5), *G*_{1 }\[\left( 2, 4 \right)\] *F*_{1}(4, 2) and *E*_{1}(4, 0).

The values of Z at these corner points are as follows

Corner point | Z= 2x + 3y |

O |
0 |

B_{1} |
15 |

G_{1} |
16 |

F_{1} |
14 |

E_{1} |
8 |

The maximum value of Z is 16 which is attained at

*G*

_{1 }\[\left( 2, 4 \right)\] Thus, the maximum profit is 16 monetary units obtained when 2 units of first product and 4 units of second product were manufacture .