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 $x = 0,\ y = \dfrac{1}{3}$.