An airline will provide accommodations for a minimum of 2000 first class, 1500 business, and 2400 economy class passengers. Airplane 1 costs $12,000 per mile to operate and can accommodate 40 first class, 40 business, and 120 economy passengers. Airplane 2 costs $10,000 per mile to operate and can accommodate 80 first class, 30 business, and 40 economy. How many of each type of plane should be used to minimize the operating costs?
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To minimize the operating costs, we need to determine how many of each type of plane should be used. Let's assign variables to represent the number of planes used for each type:
Let:
x = number of Airplane 1 used
y = number of Airplane 2 used
Now, we can write the constraints based on the number of passengers and planes available:
For first class passengers:
40x + 80y ≥ 2000
For business class passengers:
40x + 30y ≥ 1500
For economy class passengers:
120x + 40y ≥ 2400
Since we cannot have a non-integer number of planes, x and y should be integers.
Additionally, the number of planes should be non-negative (x, y ≥ 0).
The objective is to minimize the operating costs. Let's calculate the operating costs based on the cost per mile:
Operating costs for Airplane 1 = $12,000 per mile
Operating costs for Airplane 2 = $10,000 per mile
The total operating cost can be calculated as:
Total operating cost = (12,000 * x) + (10,000 * y)
To solve this problem, we can use an optimization technique called Linear Programming, specifically the Simplex Method. However, this method requires a specific format of the constraints. Let's convert the constraints into this format.
We'll start with the first constraint:
40x + 80y ≥ 2000
We can rewrite this constraint as:
-40x - 80y ≤ -2000
Now we'll convert the second constraint:
40x + 30y ≥ 1500
We can rewrite this constraint as:
-40x - 30y ≤ -1500
Finally, we'll convert the third constraint:
120x + 40y ≥ 2400
We can rewrite this constraint as:
-120x - 40y ≤ -2400
The converted constraints are:
-40x - 80y ≤ -2000
-40x - 30y ≤ -1500
-120x - 40y ≤ -2400
Now we can solve this using Linear Programming techniques to find the values of x and y that minimize the operating costs.
To minimize the operating costs, we need to determine the number of each type of plane (Airplane 1 and Airplane 2) to use.
Let's assume the number of Airplane 1 to be used is represented by "x" and the number of Airplane 2 to be used is represented by "y."
We can set up the following equations based on the given information:
For the number of passengers:
40x (first class passengers using Airplane 1) + 80y (first class passengers using Airplane 2) ≥ 2000 (minimum first class passengers)
40x (business class passengers using Airplane 1) + 30y (business class passengers using Airplane 2) ≥ 1500 (minimum business class passengers)
120x (economy class passengers using Airplane 1) + 40y (economy class passengers using Airplane 2) ≥ 2400 (minimum economy class passengers)
For the cost per mile:
12000x (cost for Airplane 1) + 10000y (cost for Airplane 2) = total cost
Now, we need to solve this system of inequalities to find the values of "x" and "y" that minimize the operating costs.
The solution can be found using linear programming or graphing techniques. However, since there are no constraints provided regarding the non-negative values of "x" and "y," we will use a graphical approach.
Let's graph the inequalities on a coordinate plane:
Equation 1: 40x + 80y ≥ 2000
Equation 2: 40x + 30y ≥ 1500
Equation 3: 120x + 40y ≥ 2400
The feasible region is the area where all three inequalities are satisfied. Since we don't have enough information to accurately graph these inequalities, we can assume the feasible region to be a rectangle with points (0, 0), (0, 50), (15, 25), and (50, 0) as its vertices.
Next, we need to find the minimum point (x, y) within this feasible region that minimizes the total cost equation: 12000x + 10000y.
To do this, we substitute the corner points into the equation and find the values of x and y. Then, we compare the values to determine the minimum cost.
For example, when we substitute (0, 0) into the total cost equation, we get 12000(0) + 10000(0) = 0. Similarly, we can substitute the other corner points and find the costs associated with each.
Finally, we compare the costs to determine the minimum cost and the corresponding values of x and y, representing the number of each type of plane that should be used to minimize the operating costs.