b. The mwanunchi manufacture company has two plants $$ P_{1} $$ and $$ P_{2} $$ which produces three types of product namely $$ A, B $$ and $$ C $$, the first plant can manuffucture one unit of $$ A $$, four units of $$ B $$ and four units of $$ C $$ at the cost of 400 /=per hour, the second plant can manufateture three units of A, five units of B and one unit of c at the cost of $$ 600 /= $$ per hour the company have received an order of 80 units of $$ A, 230 $$ units of B and 110 units of C. how many hours should be allowed to each plant in order to satisfy the order at the least cost?

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To determine the optimal number of hours each plant ($$ P_1 $$ and $$ P_2 $$) should operate to meet the order at the least cost, we can formulate and solve a linear programming problem.

### **Given:**

- **Plant $$ P_1 $$:**
  - Produces:
    - 1 unit of $$ A $$ per hour
    - 4 units of $$ B $$ per hour
    - 4 units of $$ C $$ per hour
  - Cost: \$400 per hour

- **Plant $$ P_2 $$:**
  - Produces:
    - 3 units of $$ A $$ per hour
    - 5 units of $$ B $$ per hour
    - 1 unit of $$ C $$ per hour
  - Cost: \$600 per hour

- **Order Requirements:**
  - 80 units of $$ A $$
  - 230 units of $$ B $$
  - 110 units of $$ C $$

### **Variables:**

- Let $$ x $$ be the number of hours Plant $$ P_1 $$ operates.
- Let $$ y $$ be the number of hours Plant $$ P_2 $$ operates.

### **Objective:**

Minimize the total cost:
$$ \text{Minimize } Z = 400x + 600y $$

### **Constraints:**

1. **Product $$ A $$:**
   $$ x + 3y \geq 80 $$

2. **Product $$ B $$:**
   $$ 4x + 5y \geq 230 $$

3. **Product $$ C $$:**
   $$ 4x + y \geq 110 $$

4. **Non-negativity:**
   $$ x \geq 0, \quad y \geq 0 $$

### **Solution:**

We solve the system of equations formed by the constraints to find the optimal $$ x $$ and $$ y $$.

1. **Intersection of Constraints for Products $$ A $$ and $$ B $$:**
   $$
   \begin{cases}
   x + 3y = 80 \
   4x + 5y = 230
   \end{cases}
   $$
   Solving these, we get:
   $$
   x = \frac{290}{7} \approx 41.42857 \text{ hours}, \quad y = \frac{90}{7} \approx 12.85714 \text{ hours}
   $$

2. **Verification of Constraints:**
   $$
   \text{For Product } C:
   $$
   $$
   4\left(\frac{290}{7}\right) + \left(\frac{90}{7}\right) = \frac{1250}{7} \approx 178.57 \geq 110 \quad (\text{Satisfied})
   $$

3. **Cost Calculation:**
   $$
   Z = 400\left(\frac{290}{7}\right) + 600\left(\frac{90}{7}\right) = \frac{170000}{7} \approx 24285.71
   $$

This point $$\left(\frac{290}{7}, \frac{90}{7}\right)$$ provides the minimal cost while satisfying all production requirements.

### **Conclusion:**

To satisfy the order at the least cost, the company should allocate:

- **Plant $$ P_1 $$:** $$ \frac{290}{7} $$ hours (approximately **41 3⁄7 hours**)
- **Plant $$ P_2 $$:** $$ \frac{90}{7} $$ hours (approximately **12 6⁄7 hours**)

These allocations ensure that all product requirements are met while minimizing the total operational cost.

**Answer:**  
Allocate 41 3⁄7 hours to Plant P₁ and 12 6⁄7 hours to Plant P₂. That is,

P₁ operates for 41 3/7 hours and P₂ for 12 6/7 hours.
Allocate 41 3⁄7 hours to Plant P₁ and 12 6⁄7 hours to Plant P₂.

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