Hello Students. In this post, you will find the complete** ****NCERT Solutions for Maths Class 12 Exercise 12.1**.

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**NCERT Solutions for Maths Class
12 Exercise 12.1**

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

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

Maximize Z = 3x +
4y subject to the constraints: x + y ≤ 4, x ≥ 0, y ≥ 0.

**Solution:**

Since 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 ⇒ x/4 + 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.

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

Minimize Z = –3x
+ 4y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0

**Solution:**

The objective
function is 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.

⇒ the region x + 2y ≤ 8 lies on and
below AB.

**Maths Class 12 Ex 12.1 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

The constraints are
3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0

**Maths Class 12 Ex 12.1 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:

**Maths Class
12 Ex 12.1 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

⇒ x/10 + y/5 = 1

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

Again, 3x + y ≤ 15

Let 3x + y = 15

⇒ x/5 + 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.

**Maths Class
12 Ex 12.1 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

Again, consider x + 2y ≥ 6

Let x + 2y = 6

⇒ x/6 + y/3 = 1

Here also, (0, 0) does not contain the required half plane. The double-shaded region XABY’ is the solution set. Its corners 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:**

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

Minimize and
Maximize 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

The constraints
are x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0

**Maths Class
12 Ex 12.1 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

⇒ x/100 + 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

**Maths Class
12 Ex 12.1 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

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

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

**Solution:**

The objective
function is Z = x + y

The constraints
are x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0