Linear Programming Class 12 NCERT Exercise 12.1 Solutions & Concepts

Core Mathematical Concepts & Definitions

To solve any LPP in Exercise 12.1, you must understand three key elements. First is the Objective Function: a linear function of the form Z = ax + by that we want to maximize or minimize. Second are the Constraints: linear inequalities or equations like px + qy ≤ r or px + qy ≥ r that restrict the values of our variables, along with the non-negative restrictions x ≥ 0, y ≥ 0. Third is the Feasible Region: the common region determined by all given constraints. According to Theorem 1 (Corner Point Theorem), if a feasible region is bounded, the objective function Z always achieves its optimal (maximum or minimum) value at one of the vertices (corner points) of the feasible region. If the feasible region is unbounded, an optimal value may or may not exist; if it does, it must still occur at a corner point, but we must verify it by plotting the open half-plane ax + by > M (for maximization) or ax + by < m (for minimization).

The Step-by-Step Corner Point Method

1. Plot the Boundary Lines — Treat each inequality constraint as an equation (e.g., ax + by = c) and plot these straight lines on a Cartesian plane.
2. Identify the Feasible Region — Use a test point, usually the origin (0,0), to check which side of each line satisfies the inequality. Shade the common region satisfying all constraints, including x ≥ 0, y ≥ 0 (first quadrant).
3. Determine the Corner Points — Find the coordinates of the corner points of the shaded feasible region by solving the corresponding system of linear equations for intersecting lines.
4. Evaluate and Compare — Evaluate the objective function Z = ax + by at each corner point. Identify the maximum or minimum value from these calculated values. If the region is unbounded, perform the half-plane check.

Detailed Worked Examples

• Example 1 (Bounded Region): Maximize Z = 4x + y subject to constraints: x + y ≤ 50, 3x + y ≤ 90, x ≥ 0, y ≥ 0. Step 1: Plot lines x + y = 50 (passes through (0,50) and (50,0)) and 3x + y = 90 (passes through (0,90) and (30,0)). Step 2: Since both inequalities are '≤', the feasible region is shaded towards the origin. This forms a bounded convex polygon with vertices at O(0,0), A(30,0), B(20,30), and C(0,50). Step 3: Point B is found by solving the simultaneous equations x + y = 50 and 3x + y = 90. Subtracting the first equation from the second gives 2x = 40, so x = 20, and y = 30. Step 4: Evaluate Z at the vertices: - At O(0,0): Z = 4(0) + 0 = 0 - At A(30,0): Z = 4(30) + 0 = 120 - At B(20,30): Z = 4(20) + 30 = 110 - At C(0,50): Z = 4(0) + 50 = 50 Final Answer: The maximum value of Z is 120, which occurs at the point (30, 0).
• Example 2 (Unbounded Region): Minimize Z = 3x + 5y subject to: x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0. Step 1: Plot lines x + 3y = 3 (passes through (3,0) and (0,1)) and x + y = 2 (passes through (2,0) and (0,2)). Step 2: Since the inequalities are '≥', shade the region away from the origin. The feasible region is unbounded with corner points A(3,0), B(1.5, 0.5), and C(0,2). Step 3: Evaluate Z at the vertices: - At A(3,0): Z = 3(3) + 5(0) = 9 - At B(1.5, 0.5): Z = 3(1.5) + 5(0.5) = 7 - At C(0,2): Z = 3(0) + 5(2) = 10 Step 4: Since the region is unbounded, we must check if the open half-plane 3x + 5y < 7 has any points in common with the feasible region. Plotting 3x + 5y < 7 shows it does not share any common points with the shaded region. Final Answer: The minimum value of Z is 7, which occurs at the point (1.5, 0.5).

Avoid Common Board Exam Traps in LPP

Ensure you strictly label all axes, write down the equations of each boundary line on your graph sheet, and clearly write down the coordinates of all corner points. For unbounded feasible regions, simply finding the minimum/maximum among corner points is NOT enough to score full marks in CBSE. You MUST draw the dotted line for the inequality ax + by < m (for minimization) or ax + by > M (for maximization) and state whether the resulting open half-plane has common points with the feasible region. If there are common points, no optimal value exists!

Practice Questions with Solutions

• Q: Maximize Z = 3x + 4y subject to constraints: x + y ≤ 4, x ≥ 0, y ≥ 0. A: Step 1: Write down the boundary line equation: x + y = 4. The line passes through (4,0) and (0,4). Step 2: Test point (0,0) in x + y ≤ 4. Since 0 ≤ 4 is true, the region containing the origin is the feasible region. Step 3: The corner points of the bounded feasible region are O(0,0), A(4,0), and B(0,4). Step 4: Evaluate Z at each corner point: - At O(0,0): Z = 3(0) + 4(0) = 0 - At A(4,0): Z = 3(4) + 4(0) = 12 - At B(0,4): Z = 3(0) + 4(4) = 16 Final answer: The maximum value of Z is 16 at the point (0, 4).
• Q: Minimize Z = -3x + 4y subject to constraints: x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0. A: Step 1: Draw boundary lines: L1: x + 2y = 8 (passes through (8,0), (0,4)) and L2: 3x + 2y = 12 (passes through (4,0), (0,6)). Step 2: Find intersection point of L1 and L2: Subtracting L1 from L2 gives 2x = 4 -> x = 2. Substitute x = 2 into L1: 2 + 2y = 8 -> y = 3. Point of intersection is B(2,3). Step 3: The bounded feasible region has corner points: O(0,0), A(4,0), B(2,3), and C(0,4). Step 4: Evaluate Z at each corner point: - At O(0,0): Z = -3(0) + 4(0) = 0 - At A(4,0): Z = -3(4) + 4(0) = -12 - At B(2,3): Z = -3(2) + 4(3) = 6 - At C(0,4): Z = -3(0) + 4(4) = 16 Final answer: The minimum value of Z is -12 at the point (4, 0).
• Q: Maximize Z = 5x + 3y subject to constraints: 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0. A: Step 1: Draw boundary lines: L1: 3x + 5y = 15 (passes through (5,0), (0,3)) and L2: 5x + 2y = 10 (passes through (2,0), (0,5)). Step 2: Find intersection point: Multiply L1 by 2 and L2 by 5: 6x + 10y = 30 and 25x + 10y = 50. Subtracting gives 19x = 20 -> x = 20/19. Then y = 45/19. Intersection is B(20/19, 45/19). Step 3: Corner points of the feasible region: O(0,0), A(2,0), B(20/19, 45/19), and C(0,3). Step 4: Evaluate Z at each point: - At O(0,0): Z = 0 - At A(2,0): Z = 5(2) + 3(0) = 10 - At B(20/19, 45/19): Z = 5(20/19) + 3(45/19) = (100 + 135)/19 = 235/19 ≈ 12.37 - At C(0,3): Z = 5(0) + 3(3) = 9 Final answer: The maximum value of Z is 235/19 at the point (20/19, 45/19).
• Q: Minimize and Maximize Z = 5x + 10y subject to constraints: x + 2y ≤ 120, x + y ≥ 60, x - 2y ≥ 0, x, y ≥ 0. A: Step 1: Draw boundary lines: L1: x + 2y = 120, L2: x + y = 60, L3: x - 2y = 0. Step 2: The corner points of the bounded feasible region are A(60,0), B(120,0), C(60,30), and D(40,20). Step 3: Evaluate Z at each corner point: - At A(60,0): Z = 5(60) + 10(0) = 300 - At B(120,0): Z = 5(120) + 10(0) = 600 - At C(60,30): Z = 5(60) + 10(30) = 600 - At D(40,20): Z = 5(40) + 10(20) = 400 Final answer: The minimum value of Z is 300 at (60,0). The maximum value of Z is 600, which occurs at all points along the line segment joining (120,0) and (60,30).

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