Rbse Solutions Class 12 Maths Exercise 12.1 | Graphical Method for LPP

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Get detailed, step-by-step solutions for NCERT Class 12 Maths Exercise 12.1 (Linear Programming). Learn to solve LPPs by the Graphical Method, including steps to identify the feasible region (bounded and unbounded), determine corner points, and find the maximum or minimum value of the objective function ($Z$). Includes cases for unique and multiple optimal solutions and empty feasible regions.

This exercise requires solving Linear Programming Problems (LPP) by graphically determining the feasible region and evaluating the objective function at its corner points.

1. Maximise $Z = 3x + 4y$

Subject to: $x + y \le 4$, $x \ge 0$, $y \ge 0$.

  1. Feasible Region (FR): The lines are $x=0$, $y=0$, and $x+y=4$. The FR is the polygon formed by the intersection of these inequalities (the area below $x+y=4$ in the first quadrant).
  2. Corner Points: $O(0, 0)$, $A(4, 0)$, $B(0, 4)$.
  3. Evaluate $Z$:
    • $Z(0, 0) = 3(0) + 4(0) = 0$
    • $Z(4, 0) = 3(4) + 4(0) = 12$
    • $Z(0, 4) = 3(0) + 4(4) = 16$The maximum value of $Z$ is 16 at $(0, 4)$.

2. Minimise $Z = -3x + 4y$

Subject to: $x + 2y \le 8$, $3x + 2y \le 12$, $x \ge 0$, $y \ge 0$.

  1. Feasible Region (FR):
    • $L_1: x + 2y = 8 \implies (8, 0), (0, 4)$
    • $L_2: 3x + 2y = 12 \implies (4, 0), (0, 6)$The intersection of $L_1$ and $L_2$: $(3, 2)$. The FR is the area below both lines in the first quadrant.
  2. Corner Points: $O(0, 0)$, $A(4, 0)$, $B(3, 2)$, $C(0, 4)$.
  3. Evaluate $Z$:
    • $Z(0, 0) = 0$
    • $Z(4, 0) = -3(4) + 4(0) = -12$
    • $Z(3, 2) = -3(3) + 4(2) = -9 + 8 = -1$
    • $Z(0, 4) = -3(0) + 4(4) = 16$The minimum value of $Z$ is -12 at $(4, 0)$.

3. Maximise $Z = 5x + 3y$

Subject to: $3x + 5y \le 15$, $5x + 2y \le 10$, $x \ge 0$, $y \ge 0$.

  1. Feasible Region (FR):
    • $L_1: 3x + 5y = 15 \implies (5, 0), (0, 3)$
    • $L_2: 5x + 2y = 10 \implies (2, 0), (0, 5)$The intersection of $L_1$ and $L_2$: $(20/19, 45/19)$.
  2. Corner Points: $O(0, 0)$, $A(2, 0)$, $B(20/19, 45/19)$, $C(0, 3)$.
  3. Evaluate $Z$:
    • $Z(0, 0) = 0$
    • $Z(2, 0) = 5(2) + 3(0) = 10$
    • $Z(0, 3) = 5(0) + 3(3) = 9$
    • $Z(20/19, 45/19) = 5(20/19) + 3(45/19) = (100 + 135)/19 = 235/19 \approx 12.37$The maximum value of $Z$ is $235/19$ at $(20/19, 45/19)$.

4. Minimise $Z = 3x + 5y$

Subject to: $x + 3y \ge 3$, $x + y \ge 2$, $x, y \ge 0$.

  1. Feasible Region (FR):
    • $L_1: x + 3y = 3 \implies (3, 0), (0, 1)$
    • $L_2: x + y = 2 \implies (2, 0), (0, 2)$The FR is unbounded (the area above both lines in the first quadrant).The intersection of $L_1$ and $L_2$: $(3/2, 1/2)$.
  2. Corner Points: $A(3, 0)$, $B(3/2, 1/2)$, $C(0, 2)$.
  3. Evaluate $Z$:
    • $Z(3, 0) = 3(3) + 5(0) = 9$
    • $Z(3/2, 1/2) = 3(3/2) + 5(1/2) = 9/2 + 5/2 = 14/2 = 7$
    • $Z(0, 2) = 3(0) + 5(2) = 10$
  4. Minimality Check (Unbounded Region): Since the region is unbounded, we check if $3x + 5y < 7$ has any common points with the FR. The half-plane $3x + 5y < 7$ does not intersect the FR.The minimum value of $Z$ is 7 at $(3/2, 1/2)$.

5. Maximise $Z = 3x + 2y$

Subject to: $x + 2y \le 10$, $3x + y \le 15$, $x, y \ge 0$.

  1. Feasible Region (FR):
    • $L_1: x + 2y = 10 \implies (10, 0), (0, 5)$
    • $L_2: 3x + y = 15 \implies (5, 0), (0, 15)$The intersection of $L_1$ and $L_2$: $(4, 3)$.
  2. Corner Points: $O(0, 0)$, $A(5, 0)$, $B(4, 3)$, $C(0, 5)$.
  3. Evaluate $Z$:
    • $Z(0, 0) = 0$
    • $Z(5, 0) = 3(5) + 2(0) = 15$
    • $Z(4, 3) = 3(4) + 2(3) = 12 + 6 = 18$
    • $Z(0, 5) = 3(0) + 2(5) = 10$The maximum value of $Z$ is 18 at $(4, 3)$.

6. Minimise $Z = x + 2y$

Subject to: $2x + y \ge 3$, $x + 2y \ge 6$, $x, y \ge 0$.

Show that the minimum of $Z$ occurs at more than two points.

  1. Feasible Region (FR):
    • $L_1: 2x + y = 3 \implies (3/2, 0), (0, 3)$
    • $L_2: x + 2y = 6 \implies (6, 0), (0, 3)$The FR is unbounded. The two lines intersect at $C(0, 3)$ and $L_1$ passes through $A(3/2, 0)$, $L_2$ passes through $B(6, 0)$.
  2. Corner Points: $A(3/2, 0)$, $B(6, 0)$, $C(0, 3)$.
  3. Evaluate $Z$:
    • $Z(6, 0) = 6 + 2(0) = 6$
    • $Z(0, 3) = 0 + 2(3) = 6$
    • $Z(3/2, 0) = 3/2 + 2(0) = 1.5$ (This point is not a corner point of the unbounded FR, as it is covered by the inequality $x+2y \ge 6$).Let’s re-examine the corner points. $L_1$ and $L_2$ intersect at $(0, 3)$. $L_2$ intersects the $x$-axis at $(6, 0)$.Corner Points: $A(6, 0)$, $B(0, 3)$.The FR is the area above $L_1$ and $L_2$.
  4. Evaluate $Z$ at the actual corner points:
    • $Z(6, 0) = 6$
    • $Z(0, 3) = 6$
  5. Conclusion: Since the minimum value ($Z=6$) occurs at two corner points, $A(6, 0)$ and $B(0, 3)$, the minimum value occurs at every point on the line segment connecting $A(6, 0)$ and $B(0, 3)$. This confirms the minimum occurs at more than two points.

7. Minimise and Maximise $Z = 5x + 10y$

Subject to: $x + 2y \le 120$, $x + y \ge 60$, $x – 2y \ge 0$, $x, y \ge 0$.

  1. Feasible Region (FR):
    • $L_1: x + 2y = 120 \implies (120, 0), (0, 60)$
    • $L_2: x + y = 60 \implies (60, 0), (0, 60)$
    • $L_3: x – 2y = 0 \implies (0, 0), (40, 20)$Intersection points: $L_1 \cap L_2: (0, 60)$ (No, must be $60, 0$ and $0, 60$). Let’s solve:$L_1$ and $L_3$ intersect at $(60, 30)$.$L_2$ and $L_3$ intersect at $(40, 20)$.
  2. Corner Points: $A(60, 0)$, $B(120, 0)$, $C(60, 30)$, $D(40, 20)$.
  3. Evaluate $Z$:
    • $Z(60, 0) = 5(60) + 10(0) = 300$
    • $Z(120, 0) = 5(120) + 10(0) = 600$
    • $Z(60, 30) = 5(60) + 10(30) = 300 + 300 = 600$
    • $Z(40, 20) = 5(40) + 10(20) = 200 + 200 = 400$The minimum value of $Z$ is 300 at $(60, 0)$.The maximum value of $Z$ is 600 at $(120, 0)$ and $(60, 30)$, and thus at every point on the line segment $BC$.

8. Minimise and Maximise $Z = x + 2y$

Subject to: $x + 2y \ge 100$, $2x – y \le 0$, $2x + y \le 200$; $x, y \ge 0$.

  1. Feasible Region (FR):
    • $L_1: x + 2y = 100 \implies (100, 0), (0, 50)$
    • $L_2: 2x – y = 0 \implies (0, 0), (50, 100)$ (line passes through origin)
    • $L_3: 2x + y = 200 \implies (100, 0), (0, 200)$Intersection points: $L_1 \cap L_2$ at $(20, 40)$, $L_1 \cap L_3$ at $(100, 0)$ (No, $L_3$ is $(50, 100)$ and $L_1$ is $(0, 50)$). Let’s solve $L_1 \cap L_3$: $(60, 20)$. Wait, $L_1 \cap L_3$: $x+2y=100$, $2x+y=200$. Solve: $(100, 0)$.$L_2 \cap L_3$ at $(50, 100)$.
  2. Corner Points: $A(100, 0)$, $B(50, 100)$, $C(20, 40)$. (The origin is not in the FR).
  3. Evaluate $Z$:
    • $Z(100, 0) = 100 + 2(0) = 100$
    • $Z(50, 100) = 50 + 2(100) = 250$
    • $Z(20, 40) = 20 + 2(40) = 100$The minimum value of $Z$ is 100 at $(100, 0)$ and $(20, 40)$, and thus at every point on the line segment $AC$.The maximum value of $Z$ is 250 at $(50, 100)$.

9. Maximise $Z = -x + 2y$

Subject to: $x \ge 3$, $x + y \ge 5$, $x + 2y \ge 6$, $y \ge 0$.

  1. Feasible Region (FR):
    • $L_1: x = 3$ (vertical line)
    • $L_2: x + y = 5 \implies (5, 0), (3, 2)$
    • $L_3: x + 2y = 6 \implies (6, 0), (3, 1.5)$The FR is unbounded (to the right of $x=3$ and above the other two lines).Intersection points: $L_1 \cap L_2$ at $(3, 2)$, $L_1 \cap L_3$ at $(3, 1.5)$, $L_2 \cap L_3$ at $(4, 1)$.
  2. Corner Points: $A(6, 0)$, $B(4, 1)$, $C(3, 2)$. (The point $(3, 1.5)$ is internal or on the boundary, not a corner of the final FR).
  3. Evaluate $Z$:
    • $Z(6, 0) = -6 + 2(0) = -6$
    • $Z(4, 1) = -4 + 2(1) = -2$
    • $Z(3, 2) = -3 + 2(2) = 1$
  4. Maximality Check (Unbounded Region): We check if $-x + 2y > 1$ (or $2y > x+1$) has any common points with the FR. The half-plane $-x + 2y > 1$ does intersect the feasible region (e.g., at $(3, 3)$ where $Z=3$).The objective function $Z = -x + 2y$ has No Maximum Value.

10. Maximise $Z = x + y$

Subject to: $x – y \le -1$, $-x + y \le 0$, $x, y \ge 0$.

  1. Feasible Region (FR):
    • $L_1: x – y = -1 \implies y – x = 1 \implies (0, 1), (-1, 0)$ (Shade above $L_1$)
    • $L_2: -x + y = 0 \implies y = x \implies (0, 0), (1, 1)$ (Shade above $L_2$)
  2. Analysis: The constraint $x – y \le -1$ is $y \ge x + 1$. The constraint $-x + y \le 0$ is $y \le x$.The required region must satisfy both $y \ge x + 1$ and $y \le x$.Since $x+1 > x$ for all $x$, there is no point $(x, y)$ that satisfies $y \ge x+1$ AND $y \le x$ simultaneously.The feasible region is empty (or void).Since there is no feasible region, there is No Maximum Value of $Z$.

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