2. A factory manufactures two products each requiring the use of three machines. The machine A can be used at most 70 hours; the machine B at most 50 hours; and the machine C at most 90 hours. The first product requires 2 hours on machine A, 4 hours on machine B, and 3 hours on machine C; the second product requires 5 hours on machine A, 1 hour on machine B; and 4 hours on machine C. If the profit is 2,200 per unit for the first product and 2,700 units for the second product, how many units of each product should be manufacture to maximize profit

Question

2. A factory manufactures two products each requiring the use of three machines. The machine A can be used at most 70 hours; the machine B at most 50 hours; and the machine C at most 90 hours. The first product requires 2 hours on machine A, 4 hours on machine B, and 3 hours on machine C; the second product requires 5 hours on machine A, 1 hour on machine B; and 4 hours on machine C. If the profit is 2,200 per unit for the first product and 2,700 units for the second product, how many units of each product should be manufacture to maximize profit

Answer 1

you want to maximize p=2200x+2700y

subject to
2x+5y <= 70
4x+y <= 50
3x+4y <= 90
Now just graph the constraint lines, and evaluate p at each vertex where they intersect.

Answer 2

To maximize profit, we need to determine how many units of each product should be manufactured.

Let's define the variables:
Let x be the number of units of the first product to be manufactured.
Let y be the number of units of the second product to be manufactured.

We need to set up the constraints based on the hours each machine can be used:
Machine A: 2x + 5y ≤ 70 (machine A can be used at most 70 hours)
Machine B: 4x + y ≤ 50 (machine B can be used at most 50 hours)
Machine C: 3x + 4y ≤ 90 (machine C can be used at most 90 hours)

The objective function we want to maximize is the total profit:
Profit = 2,200x + 2,700y

To solve this problem, we can use linear programming techniques. One common method is the graphical method, which involves graphing the feasible region and finding the optimal solution.

1. Graph the constraints:
Plot the lines for each constraint on a graph.

2. Identify the feasible region:
The feasible region is the area where all the constraints are satisfied. It is usually a bounded region enclosed by the lines.

3. Find the corner points:
Identify the points where the lines intersect or coincide. These are the corner points of the feasible region.

4. Evaluate the objective function at each corner point:
Substitute the x and y values of each corner point into the objective function (Profit = 2,200x + 2,700y) to find the profit at each corner point.

5. Determine the optimal solution:
The optimal solution is the corner point that yields the maximum profit.

By following these steps, you can find the values of x and y that maximize the profit.