N and M have opened a shop selling smoothies and salads. After the market, they had 8 kg of apples and 12 kg of oranges left. The salad requires 0.4 kg of apples and 0.3 kg of oranges. To make a smoothie, 0.5 kg of apples and 1 kg of oranges are required. The price of a salad is 180 rubles, the price of a smoothie is 500 rubles. What is the maximum revenue that N and M can get?

This problem can be solved as a **linear programming** problem where we aim to maximize the revenue based on the constraints related to the available quantities of apples and oranges. Here's how to set it up:

### Define Variables:
Let:
- \( x \) be the number of salads produced.
- \( y \) be the number of smoothies produced.

### Objective Function:
The revenue from selling salads and smoothies is:
\[
R = 180x + 500y
\]
We need to maximize this function.

### Constraints:
1. The total amount of apples used for salads and smoothies should not exceed 8 kg:
\[
0.4x + 0.5y \leq 8
\]
2. The total amount of oranges used for salads and smoothies should not exceed 12 kg:
\[
0.3x + 1y \leq 12
\]
3. Non-negative constraints (they can't produce a negative number of salads or smoothies):
\[
x \geq 0, \quad y \geq 0
\]

### Solution Approach:
Now, we solve this system to find the maximum revenue by evaluating the objective function at the vertices of the feasible region (determined by the constraints).

#### Step 1: Solve for intercepts
To find the boundary points, we solve the system of inequalities.

- From \( 0.4x + 0.5y = 8 \), set \( x = 0 \):
  \[
  0.5y = 8 \quad \Rightarrow \quad y = 16
  \]
  Set \( y = 0 \):
  \[
  0.4x = 8 \quad \Rightarrow \quad x = 20
  \]
  So, one line intersects the axes at \( (20, 0) \) and \( (0, 16) \).

- From \( 0.3x + y = 12 \), set \( x = 0 \):
  \[
  y = 12
  \]
  Set \( y = 0 \):
  \[
  0.3x = 12 \quad \Rightarrow \quad x = 40
  \]
  So, the other line intersects the axes at \( (40, 0) \) and \( (0, 12) \).

#### Step 2: Solve for intersections
To find the point where the two lines intersect, solve:
\[
0.4x + 0.5y = 8 \quad \text{and} \quad 0.3x + y = 12
\]
From the second equation, solve for \( y \):
\[
y = 12 - 0.3x
\]
Substitute into the first equation:
\[
0.4x + 0.5(12 - 0.3x) = 8
\]
\[
0.4x + 6 - 0.15x = 8
\]
\[
0.25x = 2 \quad \Rightarrow \quad x = 8
\]
Substitute \( x = 8 \) into \( y = 12 - 0.3x \):
\[
y = 12 - 0.3(8) = 9.6
\]
So, the lines intersect at \( (8, 9.6) \).

#### Step 3: Evaluate the objective function
Now, evaluate the objective function at the vertices of the feasible region:

- At \( (0, 12) \): 
  \[
  R = 180(0) + 500(12) = 6000 \, \text{rubles}
  \]
- At \( (0, 16) \): 
  \[
  R = 180(0) + 500(16) = 8000 \, \text{rubles}
  \]
- At \( (20, 0) \): 
  \[
  R = 180(20) + 500(0) = 3600 \, \text{rubles}
  \]
- At \( (8, 9.6) \): 
  \[
  R = 180(8) + 500(9.6) = 1440 + 4800 = 6240 \, \text{rubles}
  \]

### Conclusion:
The maximum revenue is **8000 rubles**, which is obtained by producing 16 smoothies and 0 salads.

Would you like more details or have any questions?  
Here are 5 related questions:

1. How does linear programming help in maximizing or minimizing a problem?
2. What are some other real-world applications of linear programming?
3. Can you solve a linear programming problem graphically?
4. What happens if the constraints in the problem change?
5. How do you find the feasible region in a linear programming problem?

**Tip:** When solving for multiple variables, always check the feasibility of the solutions to ensure they satisfy all constraints.

Learn how to use linear programming to maximize revenue in a shop that sells smoothies and salads. With 8 kg of apples and 12 kg of oranges, this problem involves optimization through constraints on resources. Suitable for students in Grades 10-12, the solution explores key mathematical concepts like inequalities and optimization.

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