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.