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Textbook | NCERT |

Board | CBSE |

Category | NCERT Solutions |

Class | Class 12 |

Subject | Maths |

Chapter | Chapter 12 |

Exercise | Class 12 Maths Chapter 12 Linear Programming Exercise 12.1 |

Number of Questions Solved | 10 |

Contents

## NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1

**NCERT TEXTBOOK EXERCISES**

**Ex 12.1 Class 12 Maths Question 1.**

Maximize Z = 3x + 4y

subject to the constraints: x + y ≤ 4,x ≥ 0,y ≥ 0.

**Solution:**

As x ≥ 0, y ≥ 0, therefore we shall shade the other inequalities in the first quadrant only. Now consider x + y ≤ 4.

Let x + y = 4 => $\frac { x }{ 4 } +\frac { y }{ 4 } =1$

Thus the line has 4 and 4 as intercepts along the axes. Now (0, 0) satisfies the inequation i.e., 0 + 0 ≤ 4. Now shaded region OAB is the feasible solution.

**Ex 12.1 Class 12 Maths Question 2.**

Minimize Z = -3x+4y

subject to x + 2y ≤ 8,3x + 2y ≤ 12, x ≥ 0, y ≥ 0

**Solution:**

Objective function Z = -3x + 4y

constraints are x+2y ≤ 8,

3x + 2y ≤ 12, x ≥ 0,y ≥ 0

(i) Consider the line x+2y = 8. It pass through A (8,0) and B (0,4), putting x = 0, y = 0 in x + 2y ≤ 8,0 ≤ 8 which is true.

=> region x + 2y ≤ 8 lies on and below AB.

**Ex 12.1 Class 12 Maths Question 3.**

Maximize Z = 5x+3y

subject to 3x + 5y ≤ 15,5x + 2y ≤ 10, x≥0, y≥0

**Solution:**

The objective function is Z = 5x + 3y constraints

are 3x + 5y≤15, 5x + 2y≤10,x≥0,y≥0

**Ex 12.1 Class 12 Maths Question 4.**

Minimize Z = 3x + 5y such that x + 3y ≥ 3, x + y ≥ 2,x,y ≥ 0.

**Solution:**

For plotting the graph of x + 3y = 3, we have the following table:

**Ex 12.1 Class 12 Maths Question 5.**

Maximize Z=3x+2y subject to x+2y ≤ 10, 3x+y ≤ 15, x, y ≥ 0.

**Solution:**

Consider x + 2y ≤ 10

Let x + 2y = 10

=> $\frac { x }{ 10 } +\frac { y }{ 5 } =1$

Now (0,0) satisfies the inequation, therefore the half plane containing (0,0) is the required plane.

Again 3x+2y ≤ 15

Let 3x + y = 15

=> $\frac { x }{ 5 } +\frac { y }{ 15 } =1$

It is also satisfies by (0,0) and its required half plane contains (0,0).

Now double shaded region in the first quadrant contains the solution.

**Ex 12.1 Class 12 Maths Question 6.**

Minimize Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.

**Solution:**

Consider 2x + y ≥ 3

Let 2x + y = 3

⇒ y = 3 – 2x

(0,0) is not contained in the required half plane as (0, 0) does not satisfy the inequation 2x + y ≥ 3.

Again consider x+2y≥6

Let x + 2y = 6

=> $\frac { x }{ 6 } +\frac { y }{ 3 } =1$

Here also (0,0) does not contain the required half plane. The double-shaded region XABY’ is the solution set. Its comers are A (6,0) and B (0,3). At A, Z = 6 + 0 = 6

At B, Z = 0 + 2 × 3 = 6

We see that at both points the value of Z = 6 which is minimum. In fact at every point on the line AB makes Z=6 which is also minimum.

**Show that the minimum of z occurs at more than two points.**

**Ex 12.1 Class 12 Maths Question 7.**

Minimise and Maximise Z = 5x + 10y

subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x,y≥0

**Solution:**

The objective function is Z = 5x + 10y constraints are x + 2y≤120,x+y≥60, x-2y≥0, x,y≥0

**Ex 12.1 Class 12 Maths Question 8.**

Minimize and maximize Z = x + 2y subject to x + 2y ≥ 100,2x – y ≤ 0,2x + y ≤ 200;x,y ≥ 0.

**Solution:**

Consider x + 2y ≥ 100

Let x + 2y = 100

=> $\frac { x }{ 100 } +\frac { y }{ 50 } =1$

Now x + 2y ≥ 100 represents which does not include (0,0) as it does not made it true.

Again consider 2x – y ≤ 0

Let 2x – y = 0 or y = 2x

**Ex 12.1 Class 12 Maths Question 9.**

Maximize Z = -x + 2y, subject to the constraints: x≥3, x + y ≥ 5, x + 2y ≥ 6,y ≥ 0

**Solution:**

The objective function is Z = – x + 2y.

The constraints are x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0

**Ex 12.1 Class 12 Maths Question 10.**

Maximize Z = x + y subject to x – y≤ -1, -x + y≤0,x,y≥0

**Solution:**

Objective function Z = x + y, constraints x – y≤ -1, -x + y≤0,x,y≥0

## NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Exercise 12.1 PDF

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