Linear Programming Problem

Given:
Let $x$ = quantity (in lbs) of Food I purchased per day, and $y$ = quantity (in lbs) of Food II purchased per day.

Objective Function (Cost to be minimized):
$$Z = 0.60x + 1.00y$$

Constraints:

- Calcium requirement: $10x + 4y \geq 20$
- Protein requirement: $5x + 5y \geq 20$
- Calorie requirement: $2x + 6y \geq 13$
- Non-negativity: $x \geq 0,\ y \geq 0$

LPP Formulation:
$$\text{Minimize } Z = 0.60x + 1.00y$$
$$\text{Subject to:}$$
$$10x + 4y \geq 20$$
$$5x + 5y \geq 20$$
$$2x + 6y \geq 13$$
$$x \geq 0,\ y \geq 0$$

Given:
Let $x$ = number of units of food F1 consumed per day, and $y$ = number of units of food F2 consumed per day.

Objective Function (Cost to be minimized):
$$Z = 5x + 2.5y$$

Constraints:

- Vitamin A requirement: $2x + 4y \geq 40$
- Vitamin B requirement: $3x + 2y \geq 50$
- Non-negativity: $x \geq 0,\ y \geq 0$

LPP Formulation:
$$\text{Minimize } Z = 5x + 2.5y$$
$$\text{Subject to:}$$
$$2x + 4y \geq 40$$
$$3x + 2y \geq 50$$
$$x \geq 0,\ y \geq 0$$

*(Note: The answer key lists the first constraint as $2x+4y\geq 0$, which appears to be a misprint; the correct constraint derived from the data is $2x+4y\geq 40$.)*

Given:
Let $x$ = number of thousands of bricks transported from depot A to builder P, and $y$ = number of thousands of bricks transported from depot A to builder Q.

Since total order of P = 15 (thousand), Q = 20 (thousand), R = 15 (thousand):
- Bricks sent from A to R = $30 - x - y$ (thousands)
- Bricks sent from B to P = $15 - x$ (thousands)
- Bricks sent from B to Q = $20 - y$ (thousands)
- Bricks sent from B to R = $15 - (30 - x - y) = x + y - 15$ (thousands)

Total Transportation Cost (in Rs.):
$$Z = 40x + 20y + 30(30-x-y) + 20(15-x) + 60(20-y) + 40(x+y-15)$$
$$= 40x + 20y + 900 - 30x - 30y + 300 - 20x + 1200 - 60y + 40x + 40y - 600$$
$$= (40-30-20+40)x + (20-30-60+40)y + 1800$$
$$= 30x - 30y + 1800$$

Objective Function:
$$\text{Minimize } Z = 30x - 30y + 1800$$

Constraints:
- Stock at A: $x + y \leq 30$
- Supply to P: $x \leq 15$
- Supply to Q: $y \leq 20$
- Non-negative supply from B to R: $x + y \geq 15$
- Non-negativity: $x \geq 0,\ y \geq 0$

LPP Formulation:
$$\text{Minimize } Z = 30x - 30y + 1800$$
$$\text{Subject to:}$$
$$x + y \leq 30$$
$$x \leq 15,\quad y \leq 20$$
$$x + y \geq 15$$
$$x \geq 0,\ y \geq 0$$

Given:
Let $x$ = number of hectares allocated to crop X, and $y$ = number of hectares allocated to crop Y.

Objective Function (Profit to be maximized):
$$Z = 10500x + 9000y$$

Constraints:

- Total land: $x + y \leq 50$
- Herbicide limit: $20x + 10y \leq 800$, i.e., $2x + y \leq 80$
- Non-negativity: $x \geq 0,\ y \geq 0$

LPP Formulation:
$$\text{Maximize } Z = 10500x + 9000y$$
$$\text{Subject to:}$$
$$x + y \leq 50$$
$$2x + y \leq 80$$
$$x \geq 0,\ y \geq 0$$

Given:
Let $x$ = number of Grade I inspectors assigned, and $y$ = number of Grade II inspectors assigned.

Cost Calculation per inspector per 8-hour day:

- Grade I inspector wages per day: $5 \times 8 = \text{Rs. }40$
- Grade I pieces checked per day: $20 \times 8 = 160$ pieces; error rate = $4\%$; errors = $0.04 \times 160 = 6.4$; error cost = $6.4 \times 3 = \text{Rs. }19.20$
- Total cost per Grade I inspector per day: $40 + 19.20 = \text{Rs. }59.20$

- Grade II inspector wages per day: $4 \times 8 = \text{Rs. }32$
- Grade II pieces checked per day: $14 \times 8 = 112$ pieces; error rate = $8\%$; errors = $0.08 \times 112 = 8.96$; error cost = $8.96 \times 3 = \text{Rs. }26.88$
- Total cost per Grade II inspector per day: $32 + 26.88 = \text{Rs. }58.88$

Objective Function (Daily cost to be minimized):
$$Z = 59.20x + 58.88y$$

Constraints:

- Minimum pieces inspected: $160x + 112y \geq 1500$
- Availability of Grade I: $x \leq 10$
- Availability of Grade II: $y \leq 15$
- Non-negativity: $x \geq 0,\ y \geq 0$

LPP Formulation:
$$\text{Minimize } Z = 59.20x + 58.88y$$
$$\text{Subject to:}$$
$$160x + 112y \geq 1500$$
$$x \leq 10,\quad y \leq 15$$
$$x \geq 0,\ y \geq 0$$

Step 1: Convert inequalities to equations and find intercepts.

- Line $L_1$: $-2x + y = 1$ → passes through $(0,1)$ and $(-\frac{1}{2}, 0)$
- Line $L_2$: $x = 2$ (vertical line)
- Line $L_3$: $x + y = 3$ → passes through $(3,0)$ and $(0,3)$

Step 2: Find corner points of the feasible region.

- Intersection of $x=0$ and $y=0$: $A(0,0)$
- Intersection of $x=0$ and $-2x+y=1$: $B(0,1)$
- Intersection of $-2x+y=1$ and $x+y=3$:
Subtracting: $3x = 2 \Rightarrow x = \frac{2}{3},\ y = \frac{7}{3}$ → $C\left(\frac{2}{3}, \frac{7}{3}\right)$
- Intersection of $x=2$ and $x+y=3$: $y=1$ → $D(2,1)$
- Intersection of $x=2$ and $y=0$: $E(2,0)$

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

| Corner Point | $Z = 3x + 2y$ |
|---|---|
| $A(0,0)$ | $0$ |
| $B(0,1)$ | $2$ |
| $C\left(\frac{2}{3}, \frac{7}{3}\right)$ | $2 + \frac{14}{3} = \frac{20}{3} \approx 6.67$ |
| $D(2,1)$ | $6+2=8$ |
| $E(2,0)$ | $6$ |

Step 4: Conclusion.

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

Step 1: Find the boundary line and feasible region.

Line: $2x + 3y = 1$ passes through $\left(\frac{1}{2}, 0\right)$ and $\left(0, \frac{1}{3}\right)$.

The feasible region is above this line (since $2x+3y \geq 1$) with $x \geq 0,\ y \geq 0$.

Step 2: Find corner points.

- Intersection of $2x+3y=1$ and $x=0$: $y = \frac{1}{3}$ → $A\left(0, \frac{1}{3}\right)$
- Intersection of $2x+3y=1$ and $y=0$: $x = \frac{1}{2}$ → $B\left(\frac{1}{2}, 0\right)$

The feasible region is unbounded (extends to infinity).

Step 3: Evaluate $Z = 5x - 2y$ at corner points.

| Corner Point | $Z = 5x - 2y$ |
|---|---|
| $A\left(0, \frac{1}{3}\right)$ | $0 - \frac{2}{3} = -\frac{2}{3}$ |
| $B\left(\frac{1}{2}, 0\right)$ | $\frac{5}{2} - 0 = \frac{5}{2}$ |

Step 4: Check for unboundedness.

As $y \to \infty$ (with $x=0$), $Z = -2y \to -\infty$. However, the minimum must occur at a corner point if the feasible region is bounded in the direction of minimization. Since the region is unbounded and $Z$ can decrease without bound along the $y$-axis, we check the corner point $A$.

At $A\left(0, \frac{1}{3}\right)$, $Z = -\frac{2}{3}$.

Conclusion:

The minimum value of $Z$ is $\boxed{-\dfrac{2}{3}}$ at $x = 0,\ y = \dfrac{1}{3}$.

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.

| Corner Point | $Z = -x + 2y$ |
|---|---|
| $O(0,0)$ | $0$ |
| $A(2,0)$ | $-2$ |
| $B(6,0)$ | $-6$ |
| $C(2,4)$ | $-2+8=6$ |
| $D\left(0,\frac{10}{3}\right)$ | $\frac{20}{3} \approx 6.67$ |

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.

| Corner Point | $Z = -x + 2y$ |
|---|---|
| $A(0,1)$ | $0+2=2$ |
| $B(2,3)$ | $-2+6=4$ |
| $C(0,2)$ | $0+4=4$ |

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.

| Corner Point | $Z = 4x - 2y$ |
|---|---|
| $B(0,14)$ | $0-28=-28$ |
| $A(10,4)$ | $40-8=32$ |
| $E(12,0)$ | $48-0=48$ |
| $D\left(\frac{14}{3},0\right)$ | $\frac{56}{3}-0\approx18.67$ |
| $C(0,7)$ | $0-14=-14$ |

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.

| Corner Point | $Z = 3x + 9y$ |
|---|---|
| $O(0,0)$ | $0$ |
| $A(4,0)$ | $12$ |
| $C(0,2)$ | $18$ |

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.

| Corner Point | $Z = 3x + 2y$ |
|---|---|
| $O(0,0)$ | $0$ |
| $A(0,1)$ | $2$ |
| $B\left(\frac{2}{3}, \frac{7}{3}\right)$ | $2 + \frac{14}{3} = \frac{20}{3} \approx 6.67$ |
| $D(2,1)$ | $6+2=8$ |
| $E(2,0)$ | $6$ |

Step 4: Conclusion.

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