A small shoe company makes two styles of shoes, Oxford (x) and Loafers (y). The following constraints are given: x>=30, 10<=y<=70, x+y<=120. The profit made on the sale of the shoes is given by P=315x + 280y. Use the boundary-search method to calculate the maximum profit value.

The given constraints are:

1. $x \ge 30$
2. $10 \le y \le 70$ (which means $y \ge 10$ and $y \le 70$)
3. $x + y \le 120$

To sketch the feasible region, we plot the boundary lines and determine the region that satisfies all inequalities. The vertices of this feasible region are the intersection points of these boundary lines:

• Line 1: $x = 30$
• Line 2: $y = 10$
• Line 3: $y = 70$
• Line 4: $x + y = 120$

Let's find the vertices:

• Vertex A: Intersection of $x=30$ and $y=10$. $A = (30, 10)$
• Vertex B: Intersection of $x=30$ and $y=70$. $B = (30, 70)$
• Vertex C: Intersection of $y=70$ and $x+y=120$. Substitute $y=70$ into $x+y=120 \implies x+70=120 \implies x=50$. $C = (50, 70)$
• Vertex D: Intersection of $y=10$ and $x+y=120$. Substitute $y=10$ into $x+y=120 \implies x+10=120 \implies x=110$. $D = (110, 10)$

For part 4.1, you would plot these lines on a graph and shade the polygon formed by these vertices. The feasible region is the area bounded by the points $(30,10)$, $(30,70)$, $(50,70)$, and $(110,10)$.

The profit function is $P = 315x + 280y$. We evaluate this function at each vertex of the feasible region:

• At $A=(30, 10)$: $$P_A = 315(30) + 280(10) = 9450 + 2800 = 12250$$
• At $B=(30, 70)$: $$P_B = 315(30) + 280(70) = 9450 + 19600 = 29050$$
• At $C=(50, 70)$: $$P_C = 315(50) + 280(70) = 15750 + 19600 = 35350$$
• At $D=(110, 10)$: $$P_D = 315(110) + 280(10) = 34650 + 2800 = 37450$$

The maximum profit value is the largest of these values.
The maximum profit value is $\boxed{37450}$.

The maximum profit of 37450 occurs at vertex $D=(110, 10)$.
This means the company must sell $x=110$ Oxfords and $y=10$ Loafers to achieve maximum profit.
The company must sell $\boxed{110 \text{ Oxfords and } 10 \text{ Loafers}}$.

We need to check if the point $(x, y) = (60, 50)$ satisfies all the given constraints:

1. $x \ge 30 \implies 60 \ge 30$ (True)
2. $y \ge 10 \implies 50 \ge 10$ (True)
3. $y \le 70 \implies 50 \le 70$ (True)
4. $x + y \le 120 \implies 60 + 50 \le 120 \implies 110 \le 120$ (True)

Since all constraints are satisfied, it is feasible for the company to make 60 Oxfords and 50 Loafers.
The answer is $\boxed{\text{Yes, it is feasible}}$.
Reason: The point $(60, 50)$ satisfies all the given inequalities ($x \ge 30$, $10 \le y \le 70$, and $x+y \le 120$), meaning it lies within the feasible region.