A factory produces two types of products, A and B. Each product A requires 3 hours of labor and each product B requires 2 hours of labor. The factory has a total of 60 hours of labor available per day. If they make a profit of $50 on each product A and $40 on each product B, how many of each product should they produce to maximize their daily profit?
Let's denote the number of product A as "x" and the number of product B as "y". We have two constraints in this problem: the labor constraint and the non-negativity constraint (i.e., x and y must be greater than or equal to zero).

The labor constraint can be expressed as:
3x + 2y ≤ 60

The objective function we want to maximize is the total profit:
Profit = 50x + 40y

To solve this linear programming problem, we can graph the feasible region and find the vertices. Then, we'll evaluate the objective function at each vertex to determine which one maximizes the profit.

First, rewrite the labor constraint as an equation and solve for y:
3x + 2y = 60
y = (60 - 3x) / 2

Now, plot the line representing this equation along with the non-negativity constraints (x ≥ 0 and y ≥ 0). The feasible region will be the area below the line and above the x-axis.

The vertices of the feasible region are:
(0, 0), (0, 30), (20, 0)

Now, evaluate the objective function at each vertex:

1. Profit at (0, 0) = 50(0) + 40(0) = $0
2. Profit at (0, 30) = 50(0) + 40(30) = $1200
3. Profit at (20, 0) = 50(20) + 40(0) = $1000

The maximum profit occurs when producing 0 units of product A and 30 units of product B, resulting in a daily profit of $1200.