1. Production Planning for a Beverage Company
A beverage company has two factories located in Hanoi and Da Nang. Each factory produces drinks X, Y, and Z. The production capacity of the plants per day is as follows:
Drinks	Factory in Hanoi (bottles/day)	Factory in Da Nang (bottles/day)
X	5000	3000
Y	1200	2000
Z	3500	2500
A demand of 70,000 bottles of X, 20,000 bottles of Y, and 45,000 bottles of Z is forecasted for the month of July. The operating costs per day of the plants in Hanoi and Da Nang are $7,000 and $5,000, respectively.
Question 1: The number of days for which each factory must be run in July to minimize the operating costs in meeting demand are:
•	a. 14; 8
•	b. 13; 7
•	c. 12; 9
•	d. 15; 10

To solve this problem, we need to determine the number of days each factory should operate to minimize the total operating cost while meeting the demand for each drink.

### Let's break it down step by step:

**1. Define Variables:**
- Let $ x $ be the number of days the factory in Hanoi operates.
- Let $ y $ be the number of days the factory in Da Nang operates.

**2. Production per day for each factory:**
- **Hanoi:**
  - $ 5000x $ bottles of X
  - $ 1200x $ bottles of Y
  - $ 3500x $ bottles of Z
- **Da Nang:**
  - $ 3000y $ bottles of X
  - $ 2000y $ bottles of Y
  - $ 2500y $ bottles of Z

**3. Constraints for meeting the demand:**
- Demand for X: $ 5000x + 3000y \geq 70000 $
- Demand for Y: $ 1200x + 2000y \geq 20000 $
- Demand for Z: $ 3500x + 2500y \geq 45000 $

**4. Operating costs:**
- Hanoi: $7000 per day
- Da Nang: $5000 per day

**5. Objective:**
Minimize the total cost, which is $ 7000x + 5000y $.

### Solve the system of inequalities:

**For X:**
\[ 5000x + 3000y \geq 70000 \]
Divide by 1000:
\[ 5x + 3y \geq 70 \]

**For Y:**
\[ 1200x + 2000y \geq 20000 \]
Divide by 200:
\[ 6x + 10y \geq 100 \]
Divide by 2:
\[ 3x + 5y \geq 50 \]

**For Z:**
\[ 3500x + 2500y \geq 45000 \]
Divide by 500:
\[ 7x + 5y \geq 90 \]

### Now, let's check the options by substituting them into these inequalities:

**Option a: $ x = 14 $, $ y = 8 $**
- For X: $ 5(14) + 3(8) = 70 + 24 = 94 \geq 70 $ ✓
- For Y: $ 3(14) + 5(8) = 42 + 40 = 82 \geq 50 $ ✓
- For Z: $ 7(14) + 5(8) = 98 + 40 = 138 \geq 90 $ ✓

**Option b: $ x = 13 $, $ y = 7 $**
- For X: $ 5(13) + 3(7) = 65 + 21 = 86 \geq 70 $ ✓
- For Y: $ 3(13) + 5(7) = 39 + 35 = 74 \geq 50 $ ✓
- For Z: $ 7(13) + 5(7) = 91 + 35 = 126 \geq 90 $ ✓

**Option c: $ x = 12 $, $ y = 9 $**
- For X: $ 5(12) + 3(9) = 60 + 27 = 87 \geq 70 $ ✓
- For Y: $ 3(12) + 5(9) = 36 + 45 = 81 \geq 50 $ ✓
- For Z: $ 7(12) + 5(9) = 84 + 45 = 129 \geq 90 $ ✓

**Option d: $ x = 15 $, $ y = 10 $**
- For X: $ 5(15) + 3(10) = 75 + 30 = 105 \geq 70 $ ✓
- For Y: $ 3(15) + 5(10) = 45 + 50 = 95 \geq 50 $ ✓
- For Z: $ 7(15) + 5(10) = 105 + 50 = 155 \geq 90 $ ✓

### Now, calculate the cost for each option:

- **Option a: $ 7000(14) + 5000(8) = 98000 + 40000 = 138000 $**
- **Option b: $ 7000(13) + 5000(7) = 91000 + 35000 = 126000 $**
- **Option c: $ 7000(12) + 5000(9) = 84000 + 45000 = 129000 $**
- **Option d: $ 7000(15) + 5000(10) = 105000 + 50000 = 155000 $**

### Conclusion:
The option that meets all constraints with the minimum cost is **Option b: $ 13 $ days for Hanoi and $ 7 $ days for Da Nang**, costing $126,000.

The correct answer is **b. 13; 7**.

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Would you like more details, or do you have any further questions?

### Related Questions:
1. How would the solution change if the demand for drink Z increases by 10,000 bottles?
2. What would be the total production if both factories ran for 20 days each?
3. How would the production schedule be affected if the operating cost of the Da Nang factory increased to $6,000 per day?
4. What is the minimum number of days required to meet the demand for each drink separately in each factory?
5. If the production rate of drink Y increases by 50% in Da Nang, how does it affect the production plan?

### Tip:
When solving linear programming problems like this, it's essential to define your variables clearly and systematically check each constraint to ensure you meet the requirements.

Learn how to optimize production planning for a beverage company with two factories in Hanoi and Da Nang. Solve linear programming constraints to minimize operating costs while meeting demand for drinks X, Y, and Z. Detailed step-by-step solution provided.

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