Chapter 12 Linear Programming Ex 12.1 Solutions
Question - 1 : - Maximise Z = 3x + 4y
Subject to the constraints:
Answer - 1 : -
The feasible region determined by the constraints, x + y≤ 4, x ≥ 0, y ≥ 0, is given below
O (0, 0), A (4, 0), and B (0, 4) are the corner points of thefeasible region. The values of Z at these points are given below
| Corner point | Z = 3x + 4y | |
| O (0, 0) | 0 | |
| A (4, 0) | 12 | |
| B (0, 4) | 16 | Maximum |
Hence, the maximum value of Z is 16 at the point B (0, 4)
Question - 2 : - Minimise Z = −3x + 4y
subject to
.
Answer - 2 : -
The feasible region determined by the system of constraints,
isgiven below
O (0, 0), A (4, 0), B (2, 3) and C (0, 4) are the corner pointsof the feasible region
The values of Z at these corner points are given below
| Corner point | Z = – 3x + 4y | |
| O (0, 0) | 0 | |
| A (4, 0) | -12 | Minimum |
| B (2, 3) | 6 | |
| C (0, 4) | 16 | |
Hence, the minimum value of Z is – 12 at the point (4, 0)
Question - 3 : - Maximise Z = 5x + 3y
subject to
.
Answer - 3 : -
The feasible region determined by the system of constraints, 3x +5y ≤15, 5x +2y ≤10, x ≥0, and y ≥0, are given below
O (0, 0), A (2, 0), B (0, 3) and C (20 / 19, 45 / 19) are thecorner points of the feasible region. The values of Z at these corner pointsare given below
| Corner point | Z = 5x + 3y | |
| O (0, 0) | 0 | |
| A (2, 0) | 10 | |
| B (0, 3) | 9 | |
| C (20 / 19, 45 / 19) | 235 / 19 | Maximum |
Hence, the maximum value of Z is 235 / 19 at the point (20 / 19,45 / 19)
Question - 4 : - Minimise Z = 3x + 5y
such that
Answer - 4 : -
The feasible region determined by the system of constraints, x +3y ≥ 3, x + y ≥ 2, and x, y ≥ 0 is given below
It can be seen that the feasible region is unbounded.
The corner points of the feasible region are A (3, 0), B (3 / 2,1 / 2) and C (0, 2)
The values of Z at these corner points are given below
| Corner point | Z = 3x + 5y | |
| A (3, 0) | 9 | |
| B (3 / 2, 1 / 2) | 7 | Smallest |
| C (0, 2) | 10 | |
7 may or may not be the minimum value of Z because the feasibleregion is unbounded
For this purpose, we draw the graph of the inequality, 3x + 5y< 7 and check the resulting half plane have common points with the feasibleregion or not
Hence, it can be seen that the feasible region has no commonpoint with 3x + 5y < 7
Thus, the minimum value of Z is 7 at point B (3 / 2, 1 / 2)
Question - 5 : - Maximise Z = 3x + 2y
subject to
Answer - 5 : -
The feasible region determined by the constraints, x +2y ≤10, 3x + y ≤15, x ≥0, and y ≥0, is given below
A (5, 0), B (4, 3), C (0, 5) and D (0, 0) are the corner pointsof the feasible region.
The values of Z at these corner points are given below
| Corner point | Z = 3x + 2y | |
| A (5, 0) | 15 | |
| B (4, 3) | 18 | Maximum |
| C (0, 5) | 10 | |
Hence, the maximum value of Z is 18 at the point (4, 3)
Question - 6 : - Minimise Z = x + 2y
subject to
Answer - 6 : -
The feasible region determined by the constraints, 2x + y ≥3, x +2y ≥6, x ≥0, and y ≥0, is given below
A (6, 0) and B (0, 3) are the corner points of the feasibleregion
The values of Z at the corner points are given below
| Corner point | Z = x + 2y |
| A (6, 0) | 6 |
| B (0, 3) | 6 |
Here, the values of Z at points A and B are same. If we take anyother point such as (2, 2) on line x + 2y = 6, then Z = 6
Hence, the minimum value of Z occurs for more than 2 points.
Therefore, the value of Z is minimum at every point on the linex + 2y = 6
Question - 7 : - Minimise and Maximise Z = 5x + 10y
subject to
Answer - 7 : -
The feasible region determined by the constraints, x +2y ≤120, x + y ≥60, x −2y ≥0, x ≥0, and y ≥0, is given below
A (60, 0), B (120, 0), C (60, 30), and D (40, 20) are the cornerpoints of the feasible region. The values of Z at these corner points are given
| Corner point | Z = 5x + 10y | |
| A (60, 0) | 300 | Minimum |
| B (120, 0) | 600 | Maximum |
| C (60, 30) | 600 | Maximum |
| D (40, 20) | 400 | |
The minimum value of Z is 300 at (60, 0) and the maximum valueof Z is 600 at all the points on the line segment joining (120, 0) and (60, 30)
Question - 8 : - Minimise and Maximise Z = x + 2y
subject to
.
Answer - 8 : -
The feasible region determined by the constraints, x +2y ≥100, 2x − y ≤0, 2x + y ≤200, x ≥0, and y ≥0, is given below
A (0, 50), B (20, 40), C (50, 100) and D (0, 200) are the cornerpoints of the feasible region. The values of Z at these corner points are givenbelow
| Corner point | Z = x + 2y | |
| A (0, 50) | 100 | Minimum |
| B (20, 40) | 100 | Minimum |
| C (50, 100) | 250 | |
| D (0, 200) | 400 | Maximum |
The maximum value of Z is 400 at point (0, 200) and the minimumvalue of Z is 100 at all the points on the line segment joining the points (0,50) and (20, 40)
Question - 9 : - Maximise Z = − x + 2y, subject to the constraints:
Answer - 9 : -
The feasible region determined by the constraints, 
is given below
Here, it can be seen that the feasible region is unbounded.
The values of Z at corner points A (6, 0), B (4, 1) and C (3, 2)are given below
| Corner point | Z = – x + 2y |
| A (6, 0) | Z = – 6 |
| B (4, 1) | Z = – 2 |
| C (3, 2) | Z = 1 |
Since the feasible region is unbounded, hence, z = 1 may or maynot be the maximum value
For this purpose, we graph the inequality, – x + 2y > 1, andcheck whether the resulting half plane has points in common with the feasibleregion or not.
Here, the resulting feasible region has points in common withthe feasible region
Hence, z = 1 is not the maximum value.
Z has no maximum value.
Question - 10 : - Maximise Z = x + y, subject to
Answer - 10 : -
The region determined by the constraints, isgiven below
There is no feasible region and therefore, z has no maximumvalue.