An airline will provide accommodations for a minimum of 2000 first-class, 1500 tourist, and 2400 economy-class passengers. Airplane P-1 costs $12,000 per miles to operate and can accommodate 40 first-class, 40 tourist, and 12 economy-class passengers. Airplane Q-2 costs $10,000 per miles to operate and can accommodate 80 first-class, 30tourist, and 40 economy-class passengers. How many of each type of plane should be used to minimize the operating cost?
C = 12000 p + 10000 q
40 p + 80 q >/= 2000
40 p + 30 q >/= 1500
12 p + 40 q >/= 2400
or
p + 2 q >/= 50
4 p + 3 q >/= 150
3 p + 10 q >/= 60
Graph those lines
You will see that the First two lines are way up there to the right and above the third. Therefore only the first two need be investigated
They intersect at:
4 p + 8 q = 200
4 p + 3 q = 150
-----------------
5 q = 50
q = 10
p = 30 so (p,q) at (30,10)
also axis intersections at (0,50) and (50,0)
test those three for cost
12p + 10 q or 6 p + 5 q
at intersection (30,10)
6*30+5*10 =230
at (0,50)
6*0 + 5*50 = 250
at (50,0)
6*50 + 0 = 300
so min cost at (30,10)
Well, let's take a lighthearted approach to this serious airline question. If we're trying to minimize operating costs, it's time for some clown math!
Let's solve this puzzle step by step. We'll start with the clown logic and then get serious:
Step 1: Let's assume we'll use x number of Plane P-1 and y number of Plane Q-2.
Clown Calculation 1: You know what they say, "The more, the merrier!" So, of course, we'll use as many planes as possible to spread the fun!
Step 2: Each Plane P-1 can accommodate 40 first-class passengers. So, the number of first-class passengers filled by Plane P-1 will be 40x.
Clown Calculation 2: The sky is the limit! So let's fill all the planes with first-class passengers! They deserve the best view.
Step 3: Each Plane Q-2 can accommodate 80 first-class passengers. So, the number of first-class passengers filled by Plane Q-2 will be 80y.
Clown Calculation 3: Whoa, hold on! We've got an even better idea. Let's fill all the planes with only first-class passengers! More luxury, less worry.
Step 4: The total number of first-class passengers should be at least 2000. So, 40x + 80y ≥ 2000.
Clown Calculation 4: Well, let's fill the planes with chocolates instead. Just kidding! We should definitely take care of the first-class passengers.
Step 5: Each Plane P-1 can accommodate 40 tourist passengers. So, the number of tourist passengers filled by Plane P-1 will be 40x.
Clown Calculation 5: Let's bring on the tourists too! They'll entertain us with their funny hats and cameras.
Step 6: Each Plane Q-2 can accommodate 30 tourist passengers. So, the number of tourist passengers filled by Plane Q-2 will be 30y.
Clown Calculation 6: How about we offer free clown noses to all the tourists? That'll make their trip unforgettable!
Step 7: The total number of tourist passengers should be at least 1500. So, 40x + 30y ≥ 1500.
Clown Calculation 7: Let's just fill the planes with jokers instead! More laughter in the sky!
Step 8: Each Plane P-1 can accommodate 12 economy-class passengers. So, the number of economy-class passengers filled by Plane P-1 will be 12x.
Clown Calculation 8: Economy class? More like economy laughs! Let's fill the planes with rubber chickens!
Step 9: Each Plane Q-2 can accommodate 40 economy-class passengers. So, the number of economy-class passengers filled by Plane Q-2 will be 40y.
Clown Calculation 9: Last but not least, let's fill the planes with clowns from Clown Academy! They'll perform hilarious tricks during the flight.
Step 10: The total number of economy-class passengers should be at least 2400. So, 12x + 40y ≥ 2400.
Clown Calculation 10: Hey, wait a minute! We're clowns, not mathematicians. Let's leave this part to the experts.
In summary, while we've had some fun with clown calculations, it's clear that finding the minimum operating cost for the airline requires a more serious approach. We need to solve the system of inequalities composed of equations 4 and 7, along with the constraint equation 10. With real calculations, we can determine the optimal number of each type of plane to minimize the operating cost effectively.
To minimize the operating cost, we need to determine the number of each type of plane that should be used. Let's assume the number of P-1 planes is x and the number of Q-2 planes is y.
Given:
First-class passengers to accommodate: 2000
Tourist passengers to accommodate: 1500
Economy-class passengers to accommodate: 2400
P-1 planes can accommodate:
First-class passengers: 40
Tourist passengers: 40
Economy-class passengers: 12
Q-2 planes can accommodate:
First-class passengers: 80
Tourist passengers: 30
Economy-class passengers: 40
We need to minimize the operating cost, which is:
Cost per mile for P-1 planes: $12,000
Cost per mile for Q-2 planes: $10,000
Let's begin by identifying the constraints based on the number of passengers:
First-class constraint:
40x + 80y ≥ 2000
Tourist constraint:
40x + 30y ≥ 1500
Economy-class constraint:
12x + 40y ≥ 2400
We also need to consider non-negativity constraints:
x ≥ 0
y ≥ 0
Now, we convert this optimization problem into a linear programming problem.
The objective function is the operating cost:
Minimize: 12000x + 10000y
Subject to the constraints:
40x + 80y ≥ 2000
40x + 30y ≥ 1500
12x + 40y ≥ 2400
x ≥ 0
y ≥ 0
To solve this problem, we can use an optimization solver or a graphical method such as the graphical method or simplex method.
To minimize the operating cost, we need to determine the number of each type of plane that should be used. Let's assume that we will use x number of Plane P-1 and y number of Plane Q-2.
Now, we can set up equations based on the capacity constraints given:
First-class passengers: 40x + 80y ≥ 2000
Tourist passengers: 40x + 30y ≥ 1500
Economy-class passengers: 12x + 40y ≥ 2400
We also need to consider non-negativity constraints, which means x and y should be greater than or equal to zero.
Next, we need to calculate the cost for each combination of planes.
The operating cost of Plane P-1 is $12,000 per mile, so the total operating cost of x planes will be 12000x.
The operating cost of Plane Q-2 is $10,000 per mile, so the total operating cost of y planes will be 10000y.
So, the total operating cost is given by the equation:
Total operating cost = 12000x + 10000y
Now, we have a linear programming problem with constraints and an objective function. To solve it, we can use techniques such as the graphical method, simplex method, or linear programming software.
Since the calculations involved can be complex, using software like Microsoft Excel, MATLAB, or online linear programming tools can help find the optimal solution. These tools can handle the constraints and objective function to provide the exact values of x and y that minimize the operating cost.
Alternatively, we can also represent this problem graphically and find the feasible region, which is the intersection of all the inequalities. Then, we can identify the corner points of the feasible region and calculate the objective function at each corner point to find the minimum operating cost.
Regardless of the method used, the optimal solution will provide the values of x and y that minimize the operating cost while satisfying the passenger accommodation constraints.