Class 12 Mathematics Practice Assignments: CBSE Class 12 Mathematics Linear Programming Class Test Set 01
Review targeted academic assignments with the CBSE Class 12 Mathematics Linear Programming Class Test Set 01. Built according to official CBSE standards for the 2026-27 term, these downloadable Class 12 Mathematics worksheets support effective daily practice for Chapter 12 Linear Programming.
Download Chapter 12 Linear Programming Assignment PDF with Solutions
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Class Test Questions with Solutions
Answer:
Maximise Z = 3x + 4y, subject to the constraints
x + y ≤ 1, x ≥ 0, y ≥ 0
The shaded region shown in the figure as OAB is bounded and the coordinates of the corner points O, A and B are (0, 0), (1, 0) and (0, 1) respectively.
| Corner Points | Corresponding value of Z = 3x + 4y |
|---|---|
| (0, 0) | 0 |
| (1, 0) | 3 |
| (0, 1) | 4 → Maximum |
Answer:
Maximise Z = 11x + 7y subject to the constraints x ≤ 3, y ≤ 2, x ≥ 0, y ≥ 0.
The shaded region as shown in the figure OABC is bounded and the coordinates of the corner points are (0, 0), (3, 0), (3, 2) and (0, 2), respectively.
| Corner Points | Corresponding value of Z = 11x + 7y |
|---|---|
| (0, 0) | 0 |
| (3, 0) | 33 |
| (3, 2) | 47 → Maximum |
| (0, 2) | 14 |
Hence, Z is maximum at (3, 2) and its maximum value is 47.
Answer:
As is clear from the graph, the corner points are O, A, E and D with coordinates (0, 0), (52, 0), (44, 16) and (0, 38), respectively. Also, the given region is bounded.
Here, Z = 3x + 4y
∵ 2x + y = 104 and 2x + 4y = 152
⇒ −3y = −48
⇒ y = 16 and x = 44
| Corner Points | Corresponding value of Z |
|---|---|
| (0, 0) | 0 |
| (52, 0) | 156 |
| (44, 16) | 196 → Maximum |
| (0, 38) | 152 |
Hence, Z at (44, 16) is maximum and its maximum value is 196.
Answer:
From the figure, it is clear that the feasible region is bounded with the coordinates of the corner points as (0, 3), (3, 2) and (0, 5). Here, Z = 11x + 7y.
x + 3y = 9 and x + y = 5
⇒ 2y = 4
⇒ y = 2 and x = 3
So, the intersection point of x + y = 5 and x + 3y = 9 is (3, 2).
| Corner Points | Corresponding value of Z |
|---|---|
| (0, 3) | 21 ← Minimum |
| (3, 2) | 47 |
| (0, 5) | 35 |
Hence, the minimum value of Z is 21 at (0, 3).
Answer:
| Corner Points | Corresponding value of F = 4x + 6y |
|---|---|
| (0, 2) | 12 ← Minimum |
| (3, 0) | 12 ← Minimum |
| (6, 0) | 24 |
| (6, 8) | 72 ← Maximum |
| (0, 5) | 30 |
Hence, the minimum value of F is 12 and the maximum value of F is 72.
∴ Maximum value of F − Minimum value of F = 60
Question 6. Minimise Z = 3x + 2y subject to the constraints:
x + y ≥ 8, 3x + 5y ≤ 15, x ≥ 0, y ≥ 0 [1]
Answer:
Given Z = 3x + 2y
subject to the constraints:
x + y ≥ 8
3x + 5y ≤ 15
x ≥ 0, y ≥ 0
The shaded region in the figure is as shown.
Since there is no common area, there is no minimum value.
\( x + 2y \leq 12,\ 2x + y \leq 12,\ x + \frac{5}{4}y \geq 5,\ x \geq 0,\ y \geq 0 \) [3]
Answer:
On plotting the graph of the constraints, we get the feasible region ABCDEA, where the corner points are A(5, 0), B(6, 0), C(4, 4), D(0, 6) and E(0, 4).
The values of Z at these points are as follows:
| Corner Points | Z = 600x + 400y |
|---|---|
| A(5, 0) | Z = 600 × 5 + 0 = 3000 |
| B(6, 0) | Z = 600 × 6 + 0 = 3600 |
| C(4, 4) | Z = 600 × 4 + 400 × 4 = 4000 (Maximum) |
| D(0, 6) | Z = 0 + 400 × 6 = 2400 |
| E(0, 4) | Z = 0 + 400 × 4 = 1600 |
Here, we see that at the point C(4, 4), we get the maximum value of Z, which is 4000.
Minimise and maximise Z = 3x + 9y subject to the constraints:
x + 3y ≤ 60, x + y ≥ 10, x ≤ y, x ≥ 0, y ≥ 0 [3]
Answer:
Hence, the region is bounded and the corner points of the feasible region are A(0, 20), B(0, 10), C(5, 5) and D(15, 15). Now, the values of the objective function at these points are given below:
| Corner Points | Z = 3x + 9y |
|---|---|
| A(0, 20) | Z = 3(0) + 9(20) = 180 |
| B(0, 10) | Z = 3(0) + 9(10) = 90 |
| C(5, 5) | Z = 3(5) + 9(5) = 60 |
| D(15, 15) | Z = 3(15) + 9(15) = 180 |
Here, the minimum value of Z is 60 which occurs at C(5, 5) and the maximum value of Z is 180 which occurs at two points A(0, 20) and D(15, 15). So, each point on the line segment AD will give the maximum value of Z.
x + y ≤ 8, 3x + 5y ≥ 15, x ≥ 0, y ≥ 0 [3]
Answer:
On shading the common region, we get the feasible region ABCD, whose corner points are A(5, 0), B(8, 0), C(0, 8) and D(0, 3).
The values of Z at these points are as follows:
| Corner Points | Z = 3x + 2y |
|---|---|
| A(5, 0) | Z = 3(5) + 2(0) = 15 |
| B(8, 0) | Z = 3(8) + 2(0) = 24 |
| C(0, 8) | Z = 3(0) + 2(8) = 16 |
| D(0, 3) | Z = 3(0) + 2(3) = 6 |
Hence, the minimum value of Z is 6 at D(0, 3).
9x + 12y ≤ 180, x + 3y ≤ 30, x ≥ 0, y ≥ 0 [3]
Answer:
The graphical representation of the lines is given below.
∴ The feasible region is OABC.
The corner points of the feasible region are O(0, 0), A(20, 0), B(12, 6) and C(0, 10).
| Corner Points | Z = 80x + 120y |
|---|---|
| O(0, 0) | Z = 80(0) + 120(0) = 0 |
| A(20, 0) | Z = 80(20) + 120(0) = 1600 |
| B(12, 6) | Z = 80(12) + 120(6) = 1680 (Maximum) |
| C(0, 10) | Z = 80(0) + 120(10) = 1200 |
Hence, the maximum value of Z is 1680.
1000x + 1200y ≤ 7600, 12x + 8y ≤ 72, x ≥ 0, y ≥ 0 [3]
Answer:
The graphical representation of these lines is given below.
The shaded region in the graph represents the feasible region and its extreme points are O(0, 0), A(6, 0), B(4, 3) and \( C\left(0, \frac{19}{3}\right) \).
Now, the values of Z at the extreme points are given below.
| Corner Points | Z = 50x + 40y |
|---|---|
| O(0, 0) | Z = 0 |
| A(6, 0) | Z = 300 |
| B(4, 3) | Z = 320 |
| \( C\left(0, \frac{19}{3}\right) \) | Z = 253.33 |
Hence, the maximum value of Z = 320 at (4, 3).
360x + 240y ≤ 5760, x + y ≤ 20, x ≥ 0, y ≥ 0 [3]
Answer:
The graphical representation of these lines is given below.
The shaded region in the graph represents the feasible region and its extreme points are O(0, 0), A(16, 0), B(8, 12) and C(0, 20).
Now, the values of Z at the extreme points are given below.
| Corner Points | Z = 22x + 18y |
|---|---|
| O(0, 0) | Z = 0 |
| A(16, 0) | Z = 352 |
| B(8, 12) | Z = 392 (Maximum) |
| C(0, 20) | Z = 360 |
So, Z is maximum at x = 8 and y = 12 and the maximum value of Z is 392.
Free study material for Mathematics
Chapter Assignment & Practice Material for Class 12 Mathematics Chapter 12 Linear Programming
Download Assignment: Chapter 12 Linear Programming (Class 12 Mathematics)
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