Important Questions For Class 12 Maths Chapter 12 Linear Programming are given at BYJU’S to the students who are appearing for the **board examination of CBSE-2020**. All the concepts of Linear Programming are important for students from the examination point of view.

Students can refer to all the concepts of Class 12 provided at BYJU’S for better preparation for their exams. Also, they can get the solutions for all **NCERT **questions of Class 10 Maths with detailed explanations. For important questions of 12th maths all chapters reach us at BYJU’S.

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## Important Questions & Answers For Class 12 Maths Chapter 12 Linear Programming

**Q. No. 1: Solve the following LPP graphically:**

**Maximise Z = 2x + 3y, subject to x + y ≤ 4, x ≥ 0, y ≥ 0**

**Solution:**

Let us draw the graph pf x + y =4 as below.

The shaded region (OAB) in the above figure is the feasible region determined by the system of constraints x ≥ 0, y ≥ 0 and x + y ≤ 4.

The feasible region OAB is bounded and the maximum value will occur at a corner point of the feasible region.

Corner Points are O(0, 0), A (4, 0) and B (0, 4).

Evaluate Z at each of these corner points.

Corner Point | Value of Z |

O(0, 0) | 2 (0) + 3(0) = 0 |

A (4, 0) | 2 (4) + 3(0) = 8 |

B (0, 4) | 2 (0) + 3 (4)= 12 ← maximum |

Hence, the maximum value of Z is 12 at the point (0, 4).

**Q. No. 2: ** **Solve the following linear programming problem graphically:**

**Minimise Z = 200 x + 500 y subject to the constraints:**

**x + 2y ≥ 10 **

**3x + 4y ≤ 24 **

**x ≥ 0, y ≥ 0 **

**Solution:**

Given,

Minimise Z = 200 x + 500 y … (1)

subject to the constraints:

x + 2y ≥ 10 … (2)

3x + 4y ≤ 24 … (3)

x ≥ 0, y ≥ 0 … (4)

Let us draw the graph of x + 2y = 10 and 3x + 4y = 24 as below.

The shaded region in the above figure is the feasible region ABC determined by the

system of constraints (2) to (4), which is bounded. The coordinates of corner point A, B and C are (0,5), (4,3) and (0,6) respectively.

Calculation of Z = 200x + 500y at these points.

Corner point | Value of Z |

(0, 5) | 2500 |

(4, 3) | 2300← Minimum |

(0, 6) | 3000 |

Hence, the minimum value of Z is 2300 is at the point (4, 3).

**Q. No. 3: A manufacturing company makes two types of television sets; one is black and white and the other is colour. The company has resources to make at most 300 sets a week. It takes Rs 1800 to make a black and white set and Rs 2700 to make a coloured set. The company can spend not more than Rs 648000 a week to make television sets. If it makes a profit of Rs 510 per black and white set and Rs 675 per coloured set, how many sets of each type should be produced so that the company has a maximum profit? Formulate this problem as a LPP given that the objective is to maximise the profit.**

**Solution:**

Let x and y denote, respectively, the number of black and white sets and coloured sets made each week.

Thus x ≥ 0, y ≥ 0

The company can make at most 300 sets a week, therefore, x + y ≤ 300.

Weekly cost (in Rs) of manufacturing the set is 1800x + 2700y and the company can spend up to Rs. 648000.

Therefore, 1800x + 2700y ≤ 648000

or

2x + 3y ≤ 720

The total profit on x black and white sets and y coloured sets is Rs (510x + 675y).

Let the objective function be Z = 510x + 675y.

Therefore, the mathematical formulation of the problem is as follows.

Maximise Z = 510x + 675y subject to the constraints :

x + y ≤ 300

2x + 3y ≤ 720

x ≥ 0, y ≥ 0

The graph of x + y = 30 and 2x + 3y = 720 is given below.

Corner point | Value of Z |

A(300, 0) | 153000 |

B(180, 120) | 172800 = Maximum |

C(0, 240) | 162000 |

Hence, the maximum profit will occur when 180 black & white sets and 120 coloured sets are produced.

**Q. No. 4: A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain atleast 8 units of vitamin A and 10 units of vitamin C. Food ‘I’ contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C. Food ‘II’ contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs Rs 50 per kg to purchase Food ‘I’ and Rs 70 per kg to purchase Food ‘II’. Formulate this problem as a linear programming problem to minimise the cost of such a mixture.**

**Solution:**

Let the mixture contain x kg of Food ‘I’ and y kg of Food ‘II’.

Clearly, x ≥ 0, y ≥ 0.

Tabulate the given data as below.

Resources | Food | Requirement | |

I
(x) |
II
(y) |
||

Vitamin A
(units/kg) |
2 | 1 | 8 |

Vitamin C
(units/kg) |
1 | 2 | 10 |

Cost (Rs/kg) | 50 | 70 |

Given that, the mixture must contain at least 8 units of vitamin A and 10 units of vitamin C.

Thus, the constraints are:

2x + y ≥ 8

x + 2y ≥ 10

Total cost Z of purchasing x kg of food ‘I’ and y kg of Food ‘II’ is Z = 50x + 70y

Hence, the mathematical formulation of the problem is:

Minimise Z = 50x + 70y … (1)

subject to the constraints:

2x + y ≥ 8 … (2)

x + 2y ≥ 10 … (3)

x, y ≥ 0 … (4)

Let us draw the graph of 2x + y = 8 and x + 2y = 10 as given below.

Here, observe that the feasible region is unbounded.

Let us evaluate the value of Z at the corner points A(0,8), B(2,4) and C(10,0).

Corner point | Value of Z |

A(0, 8) | 560 |

B(2, 4) | 380 = Minimum |

C(10, 0) | 500 |

Therefore, the minimum value of Z is 380 obtained at the point (2, 4).

Hence, the optimal mixing strategy for the dietician would be to mix 2 kg of Food ‘I’ and 4 kg of Food ‘II’, and with this strategy, the minimum cost of the mixture will be Rs 380.

### Practice Questions For Class 12 Maths Chapter 12 Linear Programming

- A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsman’s time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftsman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time.

(i) What number of rackets and bats must be made if the factory is to work at full capacity?

(ii) If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the maximum profit of the factory when it works at full capacity.

- A man rides his motorcycle at the speed of 50 km/hour. He has to spend Rs 2 per km on petrol. If he rides it at a faster speed of 80 km/hour, the petrol cost increases to Rs 3 per km. He has atmost Rs 120 to spend on petrol and one hour’s time. He wishes to find the maximum distance that he can travel. Express this problem as a linear programming problem.
- Two godowns A and B have a grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table:

Transportation cost per quintal (in Rs) | ||

From/To | A | B |

D | 6 | 4 |

E | 3 | 2 |

F | 2.50 | 3 |

How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?

- Minimise and Maximise Z = 5x + 10 y subject to:

x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0. - The corner points of the feasible region determined by the system of linear constraints are (0, 10), (5, 5), (15, 15), (0, 20).Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both the points (15, 15) and (0, 20) is

(A) p = q (B) p = 2q (C) q = 2p (D) q = 3p