Objective: To solve a real-world problem using linear inequalities.
Linear inequalities are especially useful in situations that require:
– multiple considerations or requirements
– more than one solution that satisfies the considerations or requirements
B. Background:
Your team has been approached by the Army to provide a solution to pack
combat rations for its soldiers. In each ration pack, there can be three types of
food items – X, Y and Z.
The table below contains the nutritional value and cost of each food item used in
the ration packs.
Nutritional Value (IU) Cost Per Piece Minimum Quantity
Food Item X 28 International Units $1.50 2
Food Item Y 25 International Units $1.30 2
Food Item Z 12 International Units $0.60 1
The Army has stipulated the following requirements:
a. Each ration pack must contain exactly 12 food items.
b. The nutritional value of each ration pack must contain at least 250 IUs.
c. The cost of each ration pack must not exceed $16.
d. There is a minimum quantity of 2 X, 2 Y and 1 Z in each ration pack.
C. Instruction:
Your team will analyse the information given above and come up with a solution
that fulfils all these requirements. The team will subsequently tweak the
solution to optimise different efficiencies.
Use the following variables in your working:
The number of food item X in a ration pack
The number of food item Y in a ration pack
The number of food item Z in a ration pack



### Step-by-Step Solution

Let's go through the steps to solve the given problem.

### Q1. Expressing Requirements Algebraically

Let:
- $ x $ be the number of food item X in a ration pack.
- $ y $ be the number of food item Y in a ration pack.
- $ z $ be the number of food item Z in a ration pack.

The requirements can be expressed as follows:

1. **Total number of food items:**
   \[
   x + y + z = 12
   \]

2. **Nutritional value:**
   \[
   28x + 25y + 12z \geq 250
   \]

3. **Cost constraint:**
   \[
   1.50x + 1.30y + 0.60z \leq 16
   \]

4. **Minimum quantity requirements:**
   \[
   x \geq 2, \quad y \geq 2, \quad z \geq 1
   \]

### Q2. Simplify Expressions Using Substitution

From the total number of food items equation:
\[
z = 12 - x - y
\]

Substitute $ z = 12 - x - y $ into the nutritional value and cost constraints.

**Nutritional value:**
\[
28x + 25y + 12(12 - x - y) \geq 250
\]
\[
28x + 25y + 144 - 12x - 12y \geq 250
\]
\[
16x + 13y + 144 \geq 250
\]
\[
16x + 13y \geq 106
\]

**Cost constraint:**
\[
1.50x + 1.30y + 0.60(12 - x - y) \leq 16
\]
\[
1.50x + 1.30y + 7.20 - 0.60x - 0.60y \leq 16
\]
\[
0.90x + 0.70y + 7.20 \leq 16
\]
\[
0.90x + 0.70y \leq 8.80
\]

The minimum quantity requirements remain the same:
\[
x \geq 2, \quad y \geq 2, \quad z \geq 1
\]

### Q3. Provide Two Solutions That Fulfil All Requirements

Let's find two solutions:

**Solution 1:**

Let's try $ x = 4 $, $ y = 5 $:

\[
z = 12 - 4 - 5 = 3
\]

Check the constraints:

**Nutritional value:**
\[
16(4) + 13(5) = 64 + 65 = 129 \geq 106 \quad \text{(Valid)}
\]

**Cost constraint:**
\[
0.90(4) + 0.70(5) = 3.60 + 3.50 = 7.10 \leq 8.80 \quad \text{(Valid)}
\]

Thus, $ x = 4 $, $ y = 5 $, $ z = 3 $ is a valid solution.

**Solution 2:**

Let's try $ x = 2 $, $ y = 6 $:

\[
z = 12 - 2 - 6 = 4
\]

Check the constraints:

**Nutritional value:**
\[
16(2) + 13(6) = 32 + 78 = 110 \geq 106 \quad \text{(Valid)}
\]

**Cost constraint:**
\[
0.90(2) + 0.70(6) = 1.80 + 4.20 = 6.00 \leq 8.80 \quad \text{(Valid)}
\]

Thus, $ x = 2 $, $ y = 6 $, $ z = 4 $ is a valid solution.

### Q4. Provide the Most Cost-Effective (Cheapest) Solution

From the valid solutions:

- $ x = 4 $, $ y = 5 $, $ z = 3 $: Cost = $ 1.50(4) + 1.30(5) + 0.60(3) = 6.00 + 6.50 + 1.80 = 14.30 $
- $ x = 2 $, $ y = 6 $, $ z = 4 $: Cost = $ 1.50(2) + 1.30(6) + 0.60(4) = 3.00 + 7.80 + 2.40 = 13.20 $

The cheapest solution is $ x = 2 $, $ y = 6 $, $ z = 4 $ with a cost of $13.20.

### Q5. Provide a Solution with the Highest Nutritional Value

From the valid solutions:

- $ x = 4 $, $ y = 5 $, $ z = 3 $: Nutritional value = $ 28(4) + 25(5) + 12(3) = 112 + 125 + 36 = 273 $
- $ x = 2 $, $ y = 6 $, $ z = 4 $: Nutritional value = $ 28(2) + 25(6) + 12(4) = 56 + 150 + 48 = 254 $

The highest nutritional value is obtained with $ x = 4 $, $ y = 5 $, $ z = 3 $ with 273 IUs.

### Q6. Evaluation of a Proposed Solution

Please provide the proposed solution for further analysis.

Do you want details or have any questions? Here are five numbered questions you can ask next:

1. Can you explain how to solve linear inequalities?
2. How can I graph the solution sets for these inequalities?
3. What methods can be used to find the optimal solution for linear programming problems?
4. Can you provide more examples of real-world problems solved by linear inequalities?
5. How do I interpret the results of a linear programming problem?

**Tip:** When solving inequalities, always check the feasibility of your solution against all constraints to ensure validity.

Explore a detailed solution for optimizing combat ration packs using linear inequalities. Learn how to set up and solve constraints for a packing problem, find feasible solutions, and identify the most cost-effective and nutritious options. Perfect for high school students studying algebra and linear programming.

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