Linear Programming Problem — NCERT Solutions

Step 1: Find the boundary lines and corner points.

• $L_1$: $-x + 3y = 10$
• $L_2$: $x + y = 6$ → intercepts $(6,0)$ and $(0,6)$
• $L_3$: $x - y = 2$ → intercepts $(2,0)$ and $(0,-2)$

Step 2: Find vertices of the feasible region.

• Intersection of $y=0$ and $x-y=2$: $x=2$ → $A(2,0)$
• Intersection of $y=0$ and $x+y=6$: $x=6$ → $B(6,0)$
• Intersection of $x+y=6$ and $-x+3y=10$:

Adding: $4y=16 \Rightarrow y=4,\ x=2$ → $C(2,4)$

• Intersection of $x-y=2$ and $-x+3y=10$:

Adding: $2y=12 \Rightarrow y=6,\ x=8$ — check $x+y=14>6$, not feasible.

• Intersection of $x=0$ and $x-y=2$: $y=-2$ — not feasible ($y<0$).
• Intersection of $x=0$ and $-x+3y=10$: $y=\frac{10}{3}$ → check $x+y=\frac{10}{3}<6$ ✓ and $x-y=-\frac{10}{3}<2$ ✓ → $D\left(0,\frac{10}{3}\right)$
• Intersection of $x=0$ and $x+y=6$: $y=6$ → check $-x+3y=18>10$, not feasible.
• Intersection of $x=0$ and $y=0$: check $x-y=0\leq2$ ✓, $-x+3y=0\leq10$ ✓, $x+y=0\leq6$ ✓ → $O(0,0)$

Step 3: Evaluate $Z = -x + 2y$ at corner points.

Step 4: Conclusion.

The minimum value of $Z$ is $\boxed{-6}$ at $x = 6,\ y = 0$.

(Note: The answer key states minimum $Z = -2$ at $x=2, y=0$; however, based on the constraints, $B(6,0)$ is feasible and gives $Z=-6$. The answer key answer of $x=2, y=0$ corresponds to the vertex $A$. The correct minimum is $-6$ at $(6,0)$.)

Step 1: Rewrite constraints.

• $L_1$: $-0.5x + y \leq 2$, i.e., $-x + 2y \leq 4$
• $L_2$: $x - y \leq -1$, i.e., $y \geq x + 1$

Step 2: Find corner points.

• Intersection of $x=0$ and $y = x+1$: $y=1$ → $A(0,1)$
• Intersection of $-x+2y=4$ and $y=x+1$:

$-x + 2(x+1) = 4 \Rightarrow x+2=4 \Rightarrow x=2,\ y=3$ → $B(2,3)$

• Intersection of $x=0$ and $-x+2y=4$: $y=2$ → $C(0,2)$

Step 3: Check feasibility of region.

The feasible region is bounded by $y \geq x+1$, $-x+2y \leq 4$, $x \geq 0$, $y \geq 0$.

Step 4: Evaluate $Z = -x + 2y$ at corner points.

Step 5: Check for unboundedness.

The feasible region is bounded (closed), so maximum exists.

Both $B(2,3)$ and $C(0,2)$ give $Z=4$. All points on the line segment joining $B$ and $C$ also give $Z=4$.

Conclusion:

There are multiple optimal solutions. The maximum value of $Z$ is $\boxed{4}$, achieved at $(0,2)$, $(2,3)$, and all points on the line segment joining them.

Step 1: Find the feasible region.

• $L_1$: $x - 2y \leq 1$
• $L_2$: $x + 2y \leq 6$
• $L_3$: $x - y \geq 3$, i.e., $y \leq x - 3$

Step 2: Check for feasibility.

From $L_3$: $y \leq x-3$. For $y \geq 0$, we need $x \geq 3$.

From $L_2$: $x + 2y \leq 6$. With $x \geq 3$ and $y \geq 0$: $x \leq 6$.

From $L_1$: $x - 2y \leq 1$, i.e., $y \geq \frac{x-1}{2}$.

From $L_3$: $y \leq x-3$.

So we need: $\frac{x-1}{2} \leq y \leq x-3$.

For this to have solutions: $\frac{x-1}{2} \leq x-3 \Rightarrow x-1 \leq 2x-6 \Rightarrow 5 \leq x$.

With $x \geq 5$ and $x+2y \leq 6$, $y \geq 0$: at $x=5$, $2y \leq 1 \Rightarrow y \leq 0.5$; and $y \leq x-3=2$; and $y \geq \frac{4}{2}=2$.

So $y \geq 2$ and $y \leq 0.5$ — contradiction. No feasible point exists.

Conclusion:

The feasible region is empty. The LPP has no solution (infeasible/unbounded as stated in answer key).

Step 1: Find the corner points of the feasible region.

Boundary lines:

• $L_1$: $x + y = 14$ → intercepts $(14,0)$ and $(0,14)$
• $L_2$: $2x + y = 24$ → intercepts $(12,0)$ and $(0,24)$
• $L_3$: $3x + 2y = 14$ → intercepts $\left(\frac{14}{3},0\right)$ and $(0,7)$

Intersections:

• $L_1 \cap L_2$: $x+y=14$ and $2x+y=24$ → $x=10,\ y=4$ → $A(10,4)$
• $L_1 \cap y\text{-axis}$: $(0,14)$ → $B(0,14)$
• $L_3 \cap y\text{-axis}$: $(0,7)$ → $C(0,7)$
• $L_3 \cap x\text{-axis}$: $\left(\frac{14}{3},0\right)$ → $D\left(\frac{14}{3},0\right)$
• $L_2 \cap x\text{-axis}$: $(12,0)$ → $E(12,0)$

Feasible region vertices (satisfying all constraints): $B(0,14)$, $A(10,4)$, $E(12,0)$, $D\left(\frac{14}{3},0\right)$, $C(0,7)$.

Step 2: Iso-cost method.

Assign a trial value to $Z$, say $Z=0$: line $4x-2y=0$, i.e., $y=2x$.

Move this line in the direction of decreasing $Z$ (since we minimize). The minimum occurs at the corner point where the iso-cost line last touches the feasible region.

Step 3: Evaluate $Z = 4x - 2y$ at corner points.

Step 4: Conclusion.

The minimum value of $Z$ is $\boxed{-28}$ at $B(0,14)$.

(Note: The answer key states $x=8, y=6$, Min $Z=20$. Re-checking: at $(8,6)$: $Z=32-12=20$; constraints: $8+6=14$ ✓, $16+6=22\leq24$ ✓, $24+12=36\geq14$ ✓. However, $B(0,14)$ gives $Z=-28 < 20$ and satisfies all constraints: $0+14=14$ ✓, $0+14=14\leq24$ ✓, $0+28=28\geq14$ ✓. The correct minimum is $Z=-28$ at $(0,14)$.)

Step 1: Find corner points of the feasible region.

Boundary lines:

• $L_1$: $x + 4y = 8$ → intercepts $(8,0)$ and $(0,2)$
• $L_2$: $x + 2y = 4$ → intercepts $(4,0)$ and $(0,2)$

Intersection of $L_1$ and $L_2$:
$(x+4y)-(x+2y)=8-4 \Rightarrow 2y=4 \Rightarrow y=2,\ x=0$ → $C(0,2)$

Corner points:

• $O(0,0)$
• $A(4,0)$ (from $L_2$ and $y=0$)
• $C(0,2)$ (intersection of $L_1$ and $L_2$)

(Note: $L_1$ gives $(8,0)$ but $x+2y=8>4$ at $(8,0)$, so $(8,0)$ is not feasible.)

Step 2: Iso-profit method.

Assign trial value $Z=18$: line $3x+9y=18$, i.e., $x+3y=6$.

Move this line parallel to itself in the increasing direction. The last point of the feasible region touched gives the maximum.

Step 3: Evaluate $Z = 3x + 9y$ at corner points.

Step 4: Conclusion.

The maximum value of $Z$ is $\boxed{18}$ at $x = 0,\ y = 2$.

Step 1: Find corner points of the feasible region.

Boundary lines:

• $L_1$: $-2x + y = 1$
• $L_2$: $x + y = 3$ → intercepts $(3,0)$ and $(0,3)$
• $L_3$: $x = 2$

Corner points:

• $O(0,0)$
• $A(0,1)$: intersection of $x=0$ and $L_1$
• $B\left(\frac{2}{3}, \frac{7}{3}\right)$: intersection of $L_1$ and $L_2$

($-2x+y=1$ and $x+y=3$: subtracting gives $3x=2$, $x=\frac{2}{3}$, $y=\frac{7}{3}$)

• $D(2,1)$: intersection of $L_3$ and $L_2$ ($x=2$, $y=1$)
• $E(2,0)$: intersection of $L_3$ and $y=0$

Step 2: Iso-profit method.

Assign trial value $Z=6$: line $3x+2y=6$, passes through $(2,0)$ and $(0,3)$.

Move this line parallel to itself in the increasing direction. The last corner point touched gives the maximum.

Step 3: Evaluate $Z = 3x + 2y$ at corner points.

Step 4: Conclusion.

The maximum value of $Z$ is $\boxed{8}$ at $x = 2,\ y = 1$.