A company produces soft drinks and has a contract requiring that a minimum of 80 units of 
chemical A and 60 units of chemical B go into each bottle of the drink. The chemicals are 
available in a prepared mix from two different suppliers. The supplier X1 has a mix of 4 units 
of A and 2 units of B that costs rupees 10, and the supplier X2 has a mix of 1 unit of A and 1 
unit of B that costs rupees 4. How many mixes from the company X1 and company X2 should 
the company purchase to honour contract requirement and yet minimise cost? 


This is a typical Linear Programming (LP) problem where we need to minimize the cost subject to certain constraints.

### Step 1: Define the variables
Let:
- \( x_1 \) be the number of mixes purchased from company \( X_1 \).
- \( x_2 \) be the number of mixes purchased from company \( X_2 \).

### Step 2: Formulate the objective function
The objective is to minimize the total cost. The cost function is:
\[
\text{Minimize } Z = 10x_1 + 4x_2
\]
where \( 10 \) and \( 4 \) are the costs per mix from companies \( X_1 \) and \( X_2 \) respectively.

### Step 3: Formulate the constraints
The constraints are based on the minimum requirements of chemical A and chemical B:
- For chemical A: 
  \[
  4x_1 + x_2 \geq 80
  \]
- For chemical B:
  \[
  2x_1 + x_2 \geq 60
  \]

Additionally, \( x_1 \) and \( x_2 \) must be non-negative:
\[
x_1 \geq 0, \quad x_2 \geq 0
\]

### Step 4: Solve the system of inequalities

To solve this problem, let's first graphically analyze it by finding the intersection points of the constraints. However, to make it quicker, let's proceed by solving the system of linear equations derived from setting the inequalities as equalities.

#### 1. Intersection of \( 4x_1 + x_2 = 80 \) and \( 2x_1 + x_2 = 60 \):

Subtract the second equation from the first:
\[
(4x_1 + x_2) - (2x_1 + x_2) = 80 - 60
\]
\[
2x_1 = 20 \quad \Rightarrow \quad x_1 = 10
\]

Substituting \( x_1 = 10 \) into \( 2x_1 + x_2 = 60 \):
\[
2(10) + x_2 = 60 \quad \Rightarrow \quad x_2 = 40
\]

So, one intersection point is \( (x_1, x_2) = (10, 40) \).

#### 2. Intersection with axes:
- If \( x_1 = 0 \):
  \[
  x_2 \geq 80 \quad \text{(from } 4x_1 + x_2 \geq 80)
  \]
  \[
  x_2 \geq 60 \quad \text{(from } 2x_1 + x_2 \geq 60)
  \]
  The most restrictive one is \( x_2 = 80 \).

- If \( x_2 = 0 \):
  \[
  4x_1 \geq 80 \quad \Rightarrow \quad x_1 \geq 20
  \]
  \[
  2x_1 \geq 60 \quad \Rightarrow \quad x_1 \geq 30
  \]
  The most restrictive one is \( x_1 = 30 \).

#### 3. Test all vertices:
- \( (10, 40) \):
  \[
  Z = 10(10) + 4(40) = 100 + 160 = 260
  \]
- \( (20, 0) \):
  \[
  Z = 10(20) + 4(0) = 200 + 0 = 200
  \]
- \( (30, 0) \):
  \[
  Z = 10(30) + 4(0) = 300 + 0 = 300
  \]
- \( (0, 80) \):
  \[
  Z = 10(0) + 4(80) = 0 + 320 = 320
  \]

### Step 5: Conclusion
The minimum cost is obtained at \( (x_1, x_2) = (20, 0) \) with a cost of 200 rupees.

Thus, the company should purchase 20 mixes from company \( X_1 \) and 0 mixes from company \( X_2 \) to minimize costs while meeting the contract requirements.

---

Would you like further details or have any questions?

Here are 5 related questions:
1. What would the cost be if the company decided to use only supplier \( X_2 \)?
2. How would the solution change if the cost of mix from \( X_2 \) was increased?
3. Can you explain how the intersection points were found geometrically?
4. What happens if the minimum requirements for chemical B are increased?
5. How would the solution change if the mix from \( X_1 \) only contained 3 units of A?

**Tip:** When solving linear programming problems, always check all the vertices of the feasible region to find the optimal solution.

Learn how to minimize costs using Linear Programming (LP) by optimizing supplier mixes. This problem involves balancing chemical requirements with cost constraints from two suppliers. Suitable for college-level students and professionals.

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