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The problem involves finding the optimal amounts of beef (\(x\) kg) and pork (\(y\) kg) that a family should buy to meet their daily protein and lipid needs at the lowest cost. The constraints and data are:

### Problem Statement:
- The family needs at least 900 units of protein and 400 units of lipids daily.
- 1 kg of beef contains 800 units of protein and 200 units of lipids.
- 1 kg of pork contains 600 units of protein and 400 units of lipids.
- The family can buy a maximum of 1.6 kg of beef and 1.1 kg of pork.
- The cost of 1 kg of beef is 160,000 VND, and the cost of 1 kg of pork is 110,000 VND.

### Variables:
- Let \( x \) = kg of beef bought.
- Let \( y \) = kg of pork bought.

### Constraints:
1. \( 800x + 600y \geq 900 \) (protein constraint)
2. \( 200x + 400y \geq 400 \) (lipid constraint)
3. \( 0 \leq x \leq 1.6 \) (beef constraint)
4. \( 0 \leq y \leq 1.1 \) (pork constraint)

### Objective:
Minimize the cost: \( C = 160000x + 110000y \).

### Solution:
We need to evaluate each option provided and check which one meets all the constraints at the minimum cost.

**Option A:** \( x = 0.3 \), \( y = 1.1 \)  
**Option B:** \( x = 0.3 \), \( y = 0.7 \)  
**Option C:** \( x = 0.6 \), \( y = 0.7 \)  
**Option D:** \( x = 1.6 \), \( y = 0.2 \)

Let's check each option against the constraints and compute the cost.

1. **Option A (x = 0.3, y = 1.1):**
   - Protein: \( 800 \times 0.3 + 600 \times 1.1 = 240 + 660 = 900 \) (satisfies the protein constraint)
   - Lipids: \( 200 \times 0.3 + 400 \times 1.1 = 60 + 440 = 500 \) (satisfies the lipid constraint)
   - Cost: \( 160000 \times 0.3 + 110000 \times 1.1 = 48000 + 121000 = 169000 \).

2. **Option B (x = 0.3, y = 0.7):**
   - Protein: \( 800 \times 0.3 + 600 \times 0.7 = 240 + 420 = 660 \) (does not satisfy the protein constraint)
   - Lipids: \( 200 \times 0.3 + 400 \times 0.7 = 60 + 280 = 340 \) (does not satisfy the lipid constraint).

3. **Option C (x = 0.6, y = 0.7):**
   - Protein: \( 800 \times 0.6 + 600 \times 0.7 = 480 + 420 = 900 \) (satisfies the protein constraint)
   - Lipids: \( 200 \times 0.6 + 400 \times 0.7 = 120 + 280 = 400 \) (satisfies the lipid constraint)
   - Cost: \( 160000 \times 0.6 + 110000 \times 0.7 = 96000 + 77000 = 173000 \).

4. **Option D (x = 1.6, y = 0.2):**
   - Protein: \( 800 \times 1.6 + 600 \times 0.2 = 1280 + 120 = 1400 \) (satisfies the protein constraint)
   - Lipids: \( 200 \times 1.6 + 400 \times 0.2 = 320 + 80 = 400 \) (satisfies the lipid constraint)
   - Cost: \( 160000 \times 1.6 + 110000 \times 0.2 = 256000 + 22000 = 278000 \).

### Conclusion:
The optimal choice is **Option A**: \( x = 0.3 \), \( y = 1.1 \) with the minimum cost of 169,000 VND.

Would you like to go through any details or have any further questions?

### Expanding Questions:
1. How to solve linear programming problems with multiple constraints?
2. What are the methods to minimize costs in constrained optimization problems?
3. How does changing constraints affect the optimal solution?
4. Can you explain how sensitivity analysis applies to this problem?
5. What are real-world applications of linear programming in daily life?

### Tip:
When solving optimization problems with constraints, always ensure that all the constraints are satisfied before evaluating the cost or objective function.

A family needs at least 900 units of protein and 400 units of lipids in their daily food intake. Every kilogram of beef contains 800 units of protein and 200 units of lipids, and every kilogram of pork contains 600 units of protein and 400 units of lipids. The family can buy a maximum of 1.6 kg of beef and 1.1 kg of pork. The cost per kilogram of beef is 160,000 VND, and the cost per kilogram of pork is 110,000 VND. Find the quantities of beef (x) and pork (y) to buy that meet the family's nutritional needs at the lowest cost. The options for x and y are: A. x = 0.3, y = 1.1 B. x = 0.3, y = 0.7 C. x = 0.6, y = 0.7 D. x = 1.6, y = 0.2

Learn how to determine the optimal amounts of beef and pork to buy for a family to meet their daily protein and lipid needs, while minimizing costs. This linear optimization problem is suitable for high school students learning algebra and cost minimization techniques.

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