This is a linear programming problem where we need to maximize the total profit based on the constraints of raw material availability and demand limits for solutions A and B.

Step 1: Define variables

Let:

• $x$ be the number of units of solution A produced daily.
• $y$ be the number of units of solution B produced daily.

Step 2: Objective function

We aim to maximize profit, which is given by: $\text{Profit} = 8x + 10y$ where:

• $8x$ is the profit from solution A (since each unit of A gives $8),
• $10y$ is the profit from solution B (since each unit of B gives $10).

Step 3: Constraints

The constraints come from the availability of raw materials and the demand for each solution:

1. Raw material I:

   • Each unit of A uses 0.5 unit of I.
   • Each unit of B uses 0.5 unit of I.
   • The total amount of raw material I available is 150 units. $0.5x + 0.5y \leq 150$

2. Raw material II:

   • Each unit of A uses 0.6 unit of II.
   • Each unit of B uses 0.4 unit of II.
   • The total amount of raw material II available is 145 units. $0.6x + 0.4y \leq 145$

3. Demand for solution A:

   • The demand for solution A lies between 30 and 150 units. $30 \leq x \leq 150$

4. Demand for solution B:

   • The demand for solution B lies between 40 and 200 units. $40 \leq y \leq 200$

5. Non-negativity: $x \geq 0, \quad y \geq 0$

Step 4: Solve the system of equations

We now solve this linear programming problem, which we can do using the Simplex method or graphical methods. However, it can be more efficiently solved with linear programming solvers like Excel, Python (with libraries such as scipy.optimize), or other tools. Here's how the problem looks for solving:

$\text{Maximize } 8x + 10y$ subject to: $0.5x + 0.5y \leq 150$ $0.6x + 0.4y \leq 145$ $30 \leq x \leq 150$ $40 \leq y \leq 200$ $x \geq 0, \quad y \geq 0$

Would you like to proceed with solving it computationally or would you prefer a manual graphical approach to visualize this?