LINEAR PROGRAMMING EXERCISE 12.1 SOLUTION
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LINEAR PROGRAMMING EXERCISE 12.1 SOLUTION

NCERT Solutions for Class 12 Maths Chapter 12 LINEAR PROGRAMMING EXERCISE 12.1 SOLUTION

Question 1.

To solve the linear programming problem graphically, we first plot the feasible region determined by the constraints and then identify the optimal solution within this region.

Maximize Z = 3x + 4y

Constraints:

  • x + y ā‰¤ 4
  • x ā‰„ 0
  • y ā‰„ 0

Graphical Solution Steps:

  1. Plot the feasible region:
    • Plot the line x + y = 4. To do this, find the intercepts: when x = 0, y = 4, and when y = 0, x = 4. Connect these points to form the line.
    • Since x ā‰„ 0 and y ā‰„ 0, the feasible region is the area bounded by this line and the axes (quadrant I).
  2. Identify corner points:

    3. The corner points of the feasible region are where the boundary lines intersect.

  3. Evaluate the objective function at each corner point:

    4. For each corner point, substitute the coordinates into the objective function Z = 3x + 4y.

  4. Determine the optimal solution:

    5. The optimal solution is the corner point that maximizes the objective function.

Example:

Consider the constraints:

  • x + y ā‰¤ 4
  • x ā‰„ 0
  • y ā‰„ 0

Feasible Region:

  • Plot the line x + y = 4. The intercepts are (0, 4) and (4, 0).
  • Connect these points to form the line and shade the area below it, including the axes.

Corner Points:

  • The corner points of the feasible region are (0, 0), (0, 4), and (4, 0).

Objective Function:

Evaluate Z = 3x + 4y at each corner point:

  • At (0, 0): Z = 3(0) + 4(0) = 0
  • At (0, 4): Z = 3(0) + 4(4) = 16
  • At (4, 0): Z = 3(4) + 4(0) = 12

Optimal Solution:

The maximum value of Z is achieved at (0, 4), where Z = 16.

Therefore, the optimal solution to the linear programming problem is x = 0 and y = 4, with Z = 16.

Question 2.

To solve the linear programming problem graphically, we need to minimize the objective function Z = -3x + 4y subject to the given constraints. Here's a step-by-step guide:

Formulas:

  1. Objective function: Z = -3x + 4y
  2. Constraints:
    • a. x + 2y ā‰¤ 8
    • b. 3x + 2y ā‰¤ 12
    • c. x ā‰„ 0
    • d. y ā‰„ 0

Step-by-Step Solution:

  1. Plot the Feasible Region:
    • Plot the inequalities representing each constraint on the xy-plane.
    • Shade the region that satisfies all constraints. This region is the feasible region.
  2. Find Corner Points of the Feasible Region:
    • Identify the intersection points of the lines representing the constraints.
    • These intersection points are the corner points of the feasible region.
  3. Evaluate Objective Function at Corner Points:
    • Substitute the coordinates of each corner point into the objective function Z = -3x + 4y.
    • Calculate the value of Z at each corner point.
  4. Identify Minimum Value of Z:
    • Compare the values of Z at all corner points.
    • The minimum value of Z corresponds to the corner point where Z is the lowest.

Example:

Given constraints:

  1. x + 2y ā‰¤ 8
  2. 3x + 2y ā‰¤ 12
  3. x ā‰„ 0
  4. y ā‰„ 0

Feasible Region:

Plotting the inequalities, we determine the feasible region bounded by the lines:

  1. x + 2y = 8
  2. 3x + 2y = 12
  3. x = 0
  4. y = 0

Corner Points:

Solve the system of equations, we find the corner points:

  1. (0, 0)
  2. (0, 4)
  3. (4, 0)
  4. (8/3, 4/3)

Objective Function:

Evaluate Z = -3x + 4y at each corner point:

  1. Zā‚ = -3(0) + 4(0) = 0
  2. Zā‚‚ = -3(0) + 4(4) = 16
  3. Zā‚ƒ = -3(4) + 4(0) = -12
  4. Zā‚„ = -3(8/3) + 4(4/3) ā‰ˆ -4/3

Minimum Value of Z:

The minimum value of Z is achieved -12 at the corner point (4 ,0) that minimizes the objective function.

Question 3.

To solve the linear programming problem graphically, we start by plotting the feasible region determined by the given constraints. Then, we evaluate the objective function Z = 5x + 3y at the corner points of the feasible region to find the maximum value of Z.

Formulas:

  1. Objective function: Z = 5x + 3y
  2. Constraints:
    • a. 3x + 5y ā‰¤ 15
    • b. 5x + 2y ā‰¤ 10
    • c. x ā‰„ 0
    • d. y ā‰„ 0

Step-by-Step Solution:

  1. Plot the Feasible Region:
    • For each constraint, plot the corresponding inequality on the xy-plane.
    • Shade the region that satisfies all constraints.
    • The feasible region is the overlapping shaded area.
  2. Find Corner Points of the Feasible Region:
    • The corner points are the intersection points of the lines representing the constraints.
  3. Evaluate Objective Function at Corner Points:
    • Substitute the coordinates of each corner point into the objective function Z = 5x + 3y.
    • Calculate the value of Z at each corner point.
  4. Identify Maximum Value of Z:
    • Compare the values of Z at all corner points.
    • The maximum value of Z corresponds to the corner point where Z is the highest.

Example:

Given constraints:

  1. 3x + 5y ā‰¤ 15
  2. 5x + 2y ā‰¤ 10
  3. x ā‰„ 0
  4. y ā‰„ 0

Feasible Region:

Plotting the inequalities, we get a feasible region bounded by the lines:

  1. 3x + 5y = 15
  2. 5x + 2y = 10
  3. x = 0
  4. y = 0

Corner Points:

Solve the system of equations to find the corner points:

  1. (0, 0)
  2. (0, 3)
  3. (2, 0)
  4. (10/3, 0)

Objective Function:

Evaluate Z = 5x + 3y at each corner point:

  1. Zā‚ = 5(0) + 3(0) = 0
  2. Zā‚‚ = 5(0) + 3(3) = 9
  3. Zā‚ƒ = 5(2) + 3(0) = 10
  4. Zā‚„ = 5(10/3) + 3(0) ā‰ˆ 16.67

Maximum Value of Z:

The maximum value of Z occurs at (10/3, 0) with Z ā‰ˆ 16.67.

Conclusion:

By graphically solving the linear programming problem, we determine that the maximum value of Z ā‰ˆ 16.67 is achieved at (x, y) = (10/3, 0), subject to the given constraints.

Question 4.

To solve the linear programming problem graphically, we aim to minimize the objective function Z = 3x + 5y under the given constraints. Here's a detailed approach:

Formulas:

  1. Objective function: Z = 3x + 5y
  2. Constraints:
    • a. x + 3y ā‰„ 3
    • b. x + y ā‰„ 2
    • c. x ā‰„ 0
    • d. y ā‰„ 0

Step-by-Step Solution:

  1. Plot the Feasible Region:
    • Plot the inequalities representing each constraint on the xy-plane.
    • Shade the region where all constraints are satisfied. This region is the feasible region.
  2. Find Corner Points of the Feasible Region:
    • Determine the intersection points of the lines representing the constraints.
    • These intersection points are the corner points of the feasible region.
  3. Evaluate Objective Function at Corner Points:
    • Substitute the coordinates of each corner point into the objective function Z = 3x + 5y.
    • Calculate the value of Z at each corner point.
  4. Identify Minimum Value of Z:
    • Compare the values of Z at all corner points.
    • The minimum value of Z corresponds to the corner point where Z is the lowest.

Example:

Given constraints:

  1. x + 3y ā‰„ 3
  2. x + y ā‰„ 2
  3. x ā‰„ 0
  4. y ā‰„ 0

Feasible Region:

Plotting the inequalities, we determine the feasible region bounded by the lines:

  1. x + 3y = 3
  2. x + y = 2
  3. x = 0
  4. y = 0

Corner Points:

Solve the system of equations, we find the corner points:

  1. (0, 1)
  2. (2, 0)
  3. (1, 0)
  4. (0, 1)

Objective Function:

Evaluate Z = 3x + 5y at each corner point:

  1. Zā‚ = 3(0) + 5(1) = 5
  2. Zā‚‚ = 3(2) + 5(0) = 6
  3. Zā‚ƒ = 3(1) + 5(0) = 3
  4. Zā‚„ = 3(0) + 5(1) = 5

Minimum Value of Z:

The minimum value of Z occurs at (1, 0) with Z = 3.

Question 5.

To solve the linear programming problem graphically, we seek to maximize the objective function Z = 3x + 2y under the given constraints. Here's a detailed approach:

Formulas:

  1. Objective function: Z = 3x + 2y
  2. Constraints:
    • a. x + 2y ā‰¤ 10
    • b. 3x + y ā‰¤ 15
    • c. x ā‰„ 0
    • d. y ā‰„ 0

Step-by-Step Solution:

  1. Plot the Feasible Region:
    • Plot the inequalities representing each constraint on the xy-plane.
    • Shade the region where all constraints are satisfied. This region forms the feasible region.
  2. Find Corner Points of the Feasible Region:
    • Identify the intersection points of the lines representing the constraints.
    • These intersection points are the corner points of the feasible region.
  3. Evaluate Objective Function at Corner Points:
    • Substitute the coordinates of each corner point into the objective function Z = 3x + 2y.
    • Calculate the value of Z at each corner point.
  4. Identify Maximum Value of Z:
    • Compare the values of Z at all corner points.
    • The maximum value of Z corresponds to the corner point where Z is the highest.

Example:

Given constraints:

  1. x + 2y ā‰¤ 10
  2. 3x + y ā‰¤ 15
  3. x ā‰„ 0
  4. y ā‰„ 0

Feasible Region:

Plotting the inequalities, we determine the feasible region bounded by the lines:

  1. x + 2y = 10
  2. 3x + y = 15
  3. x = 0
  4. y = 0

Corner Points:

Solving the system of equations, we find the corner points:

  1. (0, 0)
  2. (0, 5)
  3. (5, 0)
  4. (5/3, 5/3)

Objective Function:

Evaluate Z = 3x + 2y at each corner point:

  1. Zā‚ = 3(0) + 2(0) = 0
  2. Zā‚‚ = 3(0) + 2(5) = 10
  3. Zā‚ƒ = 3(5) + 2(0) = 15
  4. Zā‚„ = 3(5/3) + 2(5/3) ā‰ˆ 11.67

Maximum Value of Z:

The maximum value of Z occurs at (5, 0) with Z = 15.

Question 6.

To solve the linear programming problem graphically, we aim to minimize the objective function Z = x + 2y under the given constraints. Here's a detailed approach:

Formulas:

  1. Objective function: Z = x + 2y
  2. Constraints:
    • a. 2x + y ā‰„ 3
    • b. x + 2y ā‰„ 6
    • c. x ā‰„ 0
    • d. y ā‰„ 0

Step-by-Step Solution:

  1. Plot the Feasible Region:
    • Plot the inequalities representing each constraint on the xy-plane.
    • Shade the region where all constraints are satisfied. This region forms the feasible region.
  2. Find Corner Points of the Feasible Region:
    • Identify the intersection points of the lines representing the constraints.
    • These intersection points are the corner points of the feasible region.
  3. Evaluate Objective Function at Corner Points:
    • Substitute the coordinates of each corner point into the objective function Z = x + 2y.
    • Calculate the value of Z at each corner point.
  4. Identify Minimum Value of Z:
    • Compare the values of Z at all corner points.
    • The minimum value of Z corresponds to the corner point where Z is the lowest.

Example:

Given constraints:

  1. 2x + y ā‰„ 3
  2. x + 2y ā‰„ 6
  3. x ā‰„ 0
  4. y ā‰„ 0

Feasible Region:

Plotting the inequalities, we determine the feasible region bounded by the lines:

  1. 2x + y = 3
  2. x + 2y = 6
  3. x = 0
  4. y = 0

Corner Points:

Solving the system of equations, we find the corner points:

  1. (0, 3)
  2. (0, 3)
  3. (3, 0)
  4. (2, 2)

Objective Function:

Evaluate Z = x + 2y at each corner point:

  1. Z1 = 0 + 2(3) = 6
  2. Z2 = 0 + 2(3) = 6
  3. Z3 = 3 + 2(0) = 3
  4. Z4 = 2 + 2(2) = 6

Minimum Value of Z:

The minimum value of Z occurs at (3, 0) with Z = 3.

Conclusion:

By graphically solving the linear programming problem, we determine that the minimum value of Z = 3 is achieved at (x, y) = (3, 0), subject to the given constraints.

Question 7.

Let's first analyze the given linear programming problem:

Formulas:

  1. Objective function: Z = 5x + 10y
  2. Constraints:
    1. x + 2y ā‰¤ 120
    2. x + y ā‰„ 60
    3. x - 2y ā‰„ 0
    4. x ā‰„ 0
    5. y ā‰„ 0

Analysis:

  1. Plot the Feasible Region:
    • Plot the inequalities representing each constraint on the xy-plane.
    • Shade the region where all constraints are satisfied. This forms the feasible region.
  2. Identify Corner Points of the Feasible Region:
    • Determine the intersection points of the lines representing the constraints.
    • These intersection points are the corner points of the feasible region.
  3. Evaluate Objective Function at Corner Points:
    • Substitute the coordinates of each corner point into the objective function Z = 5x + 10y.
    • Calculate the value of Z at each corner point.
  4. Identify Minimum and Maximum Value of Z:
    • Compare the values of Z at all corner points.
    • The minimum value of Z corresponds to the corner point where Z is the lowest.
    • The maximum value of Z corresponds to the corner point where Z is the highest.

Example:

Given constraints:

  1. x + 2y ā‰¤ 120
  2. x + y ā‰„ 60
  3. x - 2y ā‰„ 0
  4. x ā‰„ 0
  5. y ā‰„ 0

Feasible Region:

  • Plotting the inequalities, we determine the feasible region bounded by the lines:
    1. x + 2y = 120
    2. x + y = 60
    3. x - 2y = 0
    4. x = 0
    5. y = 0

Corner Points:

  • Solving the system of equations, we find the corner points:
    1. (0, 0)
    2. (0, 60)
    3. (40, 40)
    4. (80, 20)
    5. (120, 0)

Objective Function:

  • Evaluate Z = 5x + 10y at each corner point:
    1. Z1 = 5(0) + 10(0) = 0
    2. Z2 = 5(0) + 10(60) = 600
    3. Z3 = 5(40) + 10(40) = 600
    4. Z4 = 5(80) + 10(20) = 800
    5. Z5 = 5(120) + 10(0) = 600

Conclusion:

By evaluating Z at each corner point, we observe that the minimum value of Z occurs at three corner points: (0, 0), (0, 60), and (120, 0), all yielding Z = 0. This demonstrates that the minimum of Z occurs at more than two points. Similarly, the maximum value of Z occurs at three corner points: (0, 60), (40, 40), and (80, 20), all yielding Z = 600. Thus, both the minimum and maximum of Z occur at multiple points within the feasible region.

Question 8.

Let's analyze the given linear programming problem:

Formulas:

  1. Objective function: Z = x + 2y
  2. Constraints:
    1. x + 2y ā‰„ 100
    2. 2x - y ā‰¤ 0
    3. 2x + y ā‰¤ 200
    4. x ā‰„ 0
    5. y ā‰„ 0

Analysis:

  1. Plot the Feasible Region:
    • Plot the inequalities representing each constraint on the xy-plane.
    • Shade the region where all constraints are satisfied. This forms the feasible region.
  2. Identify Corner Points of the Feasible Region:
    • Determine the intersection points of the lines representing the constraints.
    • These intersection points are the corner points of the feasible region.
  3. Evaluate Objective Function at Corner Points:
    • Substitute the coordinates of each corner point into the objective function Z = x + 2y.
    • Calculate the value of Z at each corner point.
  4. Identify Minimum and Maximum Value of Z:
    • Compare the values of Z at all corner points.
    • The minimum value of Z corresponds to the corner point where Z is the lowest.
    • The maximum value of Z corresponds to the corner point where Z is the highest.

Example:

Given constraints:

  1. x + 2y ā‰„ 100
  2. 2x - y ā‰¤ 0
  3. 2x + y ā‰¤ 200
  4. x ā‰„ 0
  5. y ā‰„ 0

Feasible Region:

  • Plotting the inequalities, we determine the feasible region bounded by the lines:
    1. x + 2y = 100
    2. 2x - y = 0
    3. 2x + y = 200
    4. x = 0
    5. y = 0

Corner Points:

  • Solving the system of equations, we find the corner points:
    1. (0, 50)
    2. (0, 0)
    3. (50, 0)
    4. (100, 0)
    5. (0, 100)

Objective Function:

  • Evaluate Z = x + 2y at each corner point:
    1. Z1 = 0 + 2(50) = 100
    2. Z2 = 0 + 2(0) = 0
    3. Z3 = 50 + 2(0) = 50
    4. Z4 = 100 + 2(0) = 100
    5. Z5 = 0 + 2(100) = 200

Conclusion:

By evaluating Z at each corner point, we observe that the minimum value of Z occurs at three corner points: (0, 0), (50, 0), and (100, 0), all yielding Z = 0. This demonstrates that the minimum of Z occurs at more than two points. Similarly, the maximum value of Z occurs at three corner points: (0, 0), (0, 100), and (0, 50), all yielding Z = 100. Thus, both the minimum and maximum of Z occur at multiple points within the feasible region.

Question 9.

Formulas:

  1. Objective function: Z = -x + 2y
  2. Constraints:
    1. x ā‰„ 3
    2. x + y ā‰„ 5
    3. x + 2y ā‰„ 6
    4. y ā‰„ 0

Step-by-Step Solution:

  1. Plot the Feasible Region:
    • Plot the inequalities representing each constraint on the xy-plane.
    • Shade the region where all constraints are satisfied. This forms the feasible region.
  2. Identify Corner Points of the Feasible Region:
    • Determine the intersection points of the lines representing the constraints.
    • These intersection points are the corner points of the feasible region.
  3. Evaluate Objective Function at Corner Points:
    • Substitute the coordinates of each corner point into the objective function Z = -x + 2y.
    • Calculate the value of Z at each corner point.
  4. Identify Maximum Value of Z:
    • Compare the values of Z at all corner points.
    • The maximum value of Z corresponds to the corner point where Z is the highest.

Example:

Given constraints:

  1. x ā‰„ 3
  2. x + y ā‰„ 5
  3. x + 2y ā‰„ 6
  4. y ā‰„ 0

Feasible Region:

  • Plotting the inequalities, we determine the feasible region bounded by the lines:
    1. x = 3
    2. x + y = 5
    3. x + 2y = 6
    4. y = 0

Corner Points:

  • Solving the system of equations, we find the corner points:
    1. (3, 0)
    2. (3, 2)

Objective Function:

  • Evaluate Z = -x + 2y at each corner point:
    1. Z1 = -3 + 2(0) = -3
    2. Z2 = -3 + 2(2) = 1

Maximum Value of Z:

The maximum value of Z occurs at (3, 2) with Z = 1.

Conclusion:

By graphically solving the linear programming problem, we determine that the maximum value of Z = 1 is achieved at (x, y) = (3, 2), subject to the given constraints.

Question 10.

Formulas:

  1. Objective function: Z = x + y
  2. Constraints:
    1. x - y ā‰¤ -1
    2. -x + y ā‰¤ 0
    3. x ā‰„ 0
    4. y ā‰„ 0

Analysis:

  1. Plot the Feasible Region:

    • Plot the inequalities representing each constraint on the xy-plane.
    • Shade the region where all constraints are satisfied. This delineates the feasible region.

  2. Identify Corner Points of the Feasible Region:

    • Determine the intersection points of the lines representing the constraints.
    • These intersection points are the corner points of the feasible region.

  3. Evaluate Objective Function at Corner Points:

    • Substitute the coordinates of each corner point into the objective function Z = x + y.
    • Calculate the value of Z at each corner point.

  4. Identify Minimum Value of Z:

    • Compare the values of Z at all corner points.
    • The minimum value of Z corresponds to the corner point where Z is the lowest.

Example:

Given constraints:

  1. x - y ā‰¤ -1
  2. -x + y ā‰¤ 0
  3. x ā‰„ 0
  4. y ā‰„ 0

Feasible Region:

Plotting the inequalities, we determine the feasible region bounded by the lines:

  1. x - y = -1
  2. -x + y = 0
  3. x = 0
  4. y = 0

Corner Points:

Solving the system of equations, we find the corner points:

  1. (0, 0)
  2. (0, 1)
  3. (1, 0)

Objective Function:

Evaluate Z = x + y at each corner point:

  1. Zā‚ = 0 + 0 = 0
  2. Zā‚‚ = 0 + 1 = 1
  3. Zā‚ƒ = 1 + 0 = 1

Conclusion:

By evaluating Z at each corner point, we observe that the minimum value of Z occurs at three corner points: (0, 0), (0, 1), and (1, 0), all yielding Z = 0. This demonstrates that the minimum of Z occurs at more than two points.