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.
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
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
