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**Solve the following problem :**

A factory produced two types of chemicals A and B The following table gives the units of ingredients P & Q (per kg) of Chemicals A and B as well as minimum requirements of P and Q and also cost per kg. of chemicals A and B.

Ingredients per kg. /Chemical Units | A (x) |
B (y) |
Minimum requirements in |

P | 1 | 2 | 80 |

Q | 3 | 1 | 75 |

Cost (in ₹) | 4 | 6 |

Find the number of units of chemicals A and B should be produced so as to minimize the cost.

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#### Solution

Let the factory produces ‘x’ units of chemical A and ‘y’ units of chemical B

∴ Total cost Z = 4x + 6y

This is the objective function to be minimized.

From the given information, the constraints are

x + 2y ≥ 80, 3x + y ≥ 75, x ≥ 0, y ≥ 0

∴ Given problem can be formulated as

Minimize Z = 4x + 6y

Subject to, x + 2y ≥ 80, 3x + y ≥ 75, x ≥ 0, y ≥ 0

To draw the feasible region, construct table as follows:

Inequality | x + 2y ≥ 80 | 3x + y ≥ 75 |

Corresponding equation (of line) | x + 2y = 80 | 3x + y = 75 |

Intersection of line with X-axis | (80, 0) | (25, 0) |

Intersection of line with Y-axis | (0, 40) | (0, 75) |

Region | Non-origin side | Non-origin side |

Shaded portion XABCY is the feasible region,

whose vertices are A ≡ (80, 0), B and C ≡ (0, 75)

B is the point of intersection of the lines 3x + y = 75 and x + 2y = 80

Solving the above equations, we get

B ≡ (14, 33)

Here the objective function is,

Z = 4x + 6y

∴ Z at A(80, 0) = 4(80) + 6(0) = 320

Z at B(14, 33) = 4(14) + 6(33) = 254

Z at C(0, 75) = 4(0) + 6(75) = 450

∴ Z has minimum value 254 at B(14, 33)

∴ Z is mimimum, when x = 14, y = 33.

∴ Factory should produce 14 units of chemical A and 33 units of chemical B to minimize the cost to ₹ 254.