NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 12 |

Subject |
Maths |

Chapter |
Chapter 12 |

Chapter Name |
Linear Programming |

Exercise |
Ex 12.1 |

Number of Questions Solved |
10 |

Category |
NCERT Solutions |

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

**Solve the following Linear Programming Problems graphically:**

**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 =>

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.

**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.

**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

**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:

**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

=>

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

=>

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.

**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

=>

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.**

**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

**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

=>

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

**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

**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

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