To solve this linear programming problem and find the minimum and maximum values of $P = 6x + 4y$, we need to proceed through the following steps:

Step 1: Identify the constraints

The constraints given are:

1. $3x + 5y \geq 20$
2. $3x + y \leq 16$
3. $-2x + y \leq 3$
4. $x \geq 0$
5. $y \geq 0$

Step 2: Graph the constraints

We need to plot each of these inequalities on the $xy$-plane to determine the feasible region.

1. Convert inequalities to equations to find boundary lines:

   • $3x + 5y = 20$
   • $3x + y = 16$
   • $-2x + y = 3$

2. Find intersections of these lines to identify potential vertices of the feasible region. These intersections are found by solving pairs of equations:

   • Intersection of $3x + 5y = 20$ and $3x + y = 16$.
   • Intersection of $3x + y = 16$ and $-2x + y = 3$.
   • Intersection of $3x + 5y = 20$ and $-2x + y = 3$.

3. Check for feasible points within the inequalities to form the boundary of the feasible region.

Step 3: Evaluate $P = 6x + 4y$ at the vertices

Once we identify the feasible vertices from the intersections, we substitute each vertex into $P = 6x + 4y$ to find the minimum and maximum values of $P$.

Step 4: Conclusion

After evaluating $P$ at each feasible vertex, the minimum and maximum values of $P$ among those points will give us the required solution.

I'll perform the calculations for these steps to find the exact vertices and evaluate $P$ at each one.The vertices (intersections) of the feasible region are approximately:

1. $(5, 1)$
2. $\left(\frac{13}{5}, \frac{41}{5}\right) \approx (2.6, 8.2)$
3. $\left(\frac{5}{13}, \frac{49}{13}\right) \approx (0.38, 3.77)$

Step 4: Evaluate $P = 6x + 4y$ at each vertex

Next, I'll substitute each vertex into $P = 6x + 4y$ to find the minimum and maximum values of $P$.The values of $P = 6x + 4y$ at each feasible vertex are:

1. At $(5, 1)$: $P = 34$
2. At $\left(\frac{13}{5}, \frac{41}{5}\right) \approx (2.6, 8.2)$: $P = \frac{242}{5} \approx 48.4$
3. At $\left(\frac{5}{13}, \frac{49}{13}\right) \approx (0.38, 3.77)$: $P = \frac{226}{13} \approx 17.38$

Conclusion

• Maximum value of $P$: 48.4 at $\left(\frac{13}{5}, \frac{41}{5}\right)$.
• Minimum value of $P$: 17.38 at $\left(\frac{5}{13}, \frac{49}{13}\right)$.

Would you like further details on any part of this solution?

Here are some related questions for further exploration:

1. How do we handle constraints that do not intersect within the feasible region?
2. What if we change the objective function to $P = 4x + 6y$?
3. How does adding a new constraint, such as $x + y \leq 10$, affect the feasible region?
4. Can we solve this problem graphically, and how would the solution differ?
5. How do the solutions change if one of the inequalities is strict (e.g., $3x + 5y > 20$)?

Tip: When solving linear programming problems, always check for feasibility and boundedness to ensure that solutions exist and are meaningful.