To solve this linear programming problem using the method of corners, we need to:

1. Determine the feasible region by graphing each inequality and finding where they intersect.
2. Identify the corner points (vertices) of the feasible region.
3. Substitute these corner points into the objective function $P = 6x + 4y$ to find the values of $P$ at each vertex.
4. Identify the minimum and maximum values of $P$.

Step-by-Step Solution

The inequalities 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 1: Convert the inequalities to equations and solve for intersection points.

1. For $3x + 5y = 20$ and $3x + y = 16$:

   • Subtracting the second equation from the first to eliminate $3x$: $(3x + 5y) - (3x + y) = 20 - 16$ $4y = 4 \Rightarrow y = 1$
   • Substitute $y = 1$ into $3x + y = 16$: $3x + 1 = 16 \Rightarrow 3x = 15 \Rightarrow x = 5$
   • So, one intersection point is $(5, 1)$.

2. For $3x + y = 16$ and $-2x + y = 3$:

   • Subtract the second equation from the first to eliminate $y$: $(3x + y) - (-2x + y) = 16 - 3$ $5x = 13 \Rightarrow x = \frac{13}{5} = 2.6$
   • Substitute $x = 2.6$ into $3x + y = 16$: $3(2.6) + y = 16 \Rightarrow 7.8 + y = 16 \Rightarrow y = 8.2$
   • So, another intersection point is $\left( \frac{13}{5}, \frac{41}{5} \right) \approx (2.6, 8.2)$.

3. For ( 3