Problema Solution
Wilson Airlines wants to fly 1400 members of a ski club to Colorado. The airline owns two types of planes. Type A can carry 50 passengers, requires 3 flight attendants, and costs $14,000 for the trip. Type B can carry 300 passengers, requires 4 flight attendants, and costs $90,000 for the trip. If the airline must use at least as many type A planes as type B and only has 42 flight attendants, how many of each type plane should be used to minimize the cost for the trip.
Answer provided by our tutors
let
x denote the number of type A planes used in the trip,
y denote the number of type B planes used in the trip.
This word problem gives us three constraints and one objective function in terms of the variables x and y.
We need to carry at least 1400 passengers to Colorado;
our total number of passengers carried will be
(number of planes of type A)* (number of people on a type A plane)+ (number of planes of type B)*(number of people on a type B plane)
So we obtain the constraint
x* 50 + y* 300 >=1400
Similarly, the number of flight attendants we have to use is bounded above by 42.
Total number of attendants = ( number of attendants used on type A)(number of planes of type A)+( number of attendants used on type B) (number of planes of type B) <= 42
3x + 4y <= 42
Since the airline must use at least as many type A planes as type B, we have x >= y.
Of course x, y >= 0 because negative numbers don't represent planes. We have the set of constraints
x* 50 + y* 300 >=1400
3x + 4y <= 42
x >= y
x, y >= 0
What is the objective? It costs $14,000 to fly a type A plane and $90,000 to fly a Type B, so the cost to be minimized for the trip is 14000x + 90000y.
In standard form the constraints are
y >= -x/6 + 14/3
y <= - 3x/4 + 21/2
y <= x
x, y >= 0
The line L1 is given by y = -x/6 + 14/3
The line L2 is given by y = -3x/4 + 21/2
The line L3 is given by y = x
The vertex A at the intersection of L1 and L3. By solving the system
y = -x/6 + 14/3
y = x
we find x = 4
Since A is on the line y = x, A = (4; 4).
Similarly B = (6; 6) and C = (10; 3).
We check the value of the objective at each of the three vertices to nd the minimum over the feasible set:
Vertex
A = (4; 4) cost 14000*4 + 90000*4 = 416,000
B = (6; 6) cost 14000*6 + 90000*6 = 624,000
C = (10; 3) cost 14000*10 + 90000*3 = 410,000
To minimize the total cost of the trip, the airline should fly 10 type A planes and 3 type B planes.