This problem involves solving a Linear Programming (LP) optimization problem graphically. Let's break it down:

Problem Formulation

Variables:

Let:

• $x$: number of trousers produced daily.
• $y$: number of shirts produced daily.

Objective Function:

The company's goal is to maximize profit: $\text{Maximize } Z = 40x + 15y$ where:

• $40x$: profit from trousers.
• $15y$: profit from shirts.

Constraints:

1. Knitting machine constraint: $\frac{x}{1000} + \frac{y}{3000} \leq 1$ (Or equivalently: $3x + y \leq 3000$).

2. Finishing department constraint: $\frac{x}{1500} + \frac{y}{2000} \leq 1$ (Or equivalently: $4x + 3y \leq 6000$).

3. Minimum trouser production requirement: $x \geq 400$

4. Non-negativity constraints: $x \geq 0, \, y \geq 0$

Steps to Solve Graphically:

1. Plot the constraints:

   • Rewrite them as equalities and plot the lines $3x + y = 3000$ and $4x + 3y = 6000$.
   • Shade the feasible region (bounded by the constraints).

2. Identify vertices of the feasible region:

   • Find the points of intersection of the constraint lines and axes.

3. Evaluate the objective function at each vertex:

   • Substitute each vertex into $Z = 40x + 15y$.
   • The vertex giving the highest $Z$ is the optimal solution.

Would you like me to plot the graph or calculate the solution manually? Let me know! 😊

Follow-Up Questions:

1. How are the constraints converted into linear inequalities?
2. Why is it important to evaluate the objective function at the vertices of the feasible region?
3. Can this problem be solved using other methods (e.g., simplex)?
4. How does the knitting machine and finishing department limit the production?
5. Why are trousers prioritized (higher profit per unit)?

Tip:

Graphical solutions are ideal for two-variable problems, but for more variables, methods like simplex are necessary!