Question

On June 24, 1948, the former Soviet Union blocked all land and water routes through East Germany to Berlin. A gigantic airlift was organized using American and British planes to bring food, clothing, and other supplies to the more than 2 million people in West Berlin. The Americans had two types of planes available, the C-47 Skytrain and the C-54 Skymaster. The carrying capacity was 3.5 tons for a C-47 and 10 tons for a C-54. to break the Soviet Union blockade, the Western Allies had to maximize carrying capacity, but the Americans were limited by the following restrictions: -No more than 44 planes could be used per day -Each C-47 required 4 crew members per flight and the C-54 required 5 crew members per flight, the total number of personnel available per day could not exceed 200 -The Americans only had 32 C-54’s available Find the number of C-47’s and C-54’s the Americans used to maximize their carrying capacity. Part one: a. Define the variables b. Clearly state the constraints (all inequalities) related to the feasible region c. State the objective function d. Use the Graphical method to solve" graph the solution region, showing the intercepts used. Label axes, units, points, and lines. Find and label ALL corner points. Show all necessary work e. Find the number of C-47’s and C-54’s the Americans used to maximize their carrying capacity

Answer 1

a. x = # of C-47s, y = # of C-54s

b. x + y ≤ 44; 4x +5y ≤ 200; 0 ≤ y ≤ 32; 0 ≤ x

c. maximize 3.5x +10y

d. see attached

e. 10 C-47s; 32 C-54s

Explanation:

You want the constraints and the solution that maximizes cargo capacity given that 44 planes and 200 crew could be used per day to fly C-47s with a crew of 4 and a capacity of 3.5 tons, and up to 32 C-54s with a crew of 5 and a capacity of 10 tons.

a. Variables

We can let x and y represent the numbers of C-47s and C-54s flown per day, respectively.

b. Constraints

There are three constraints: number of planes, number of crew, and availability of C-54s:

• x + y ≤ 44 . . . . . . no more than 44 planes per day
• 4x +5y ≤ 200 . . . . personnel cannot exceed 200
• y ≤ 32 . . . . . . . . . . . . only 32 C-54s are available
• x ≥ 0, y ≥ 0 . . . . the numbers of planes cannot be negative

c. Objective

The objective is to maximize the total cargo capacity of the available flights. We want to ...

maximize 3.5x +10y

d. Graph

The attachment shows a graphical solution The shaded area represents the feasible region. Its vertices are marked with (#C-47s, #C-54s). The blue line shows the objective function is maximized at point (10, 32). Its value there is ...

3.5(10) +10(32) = 355 . . . . tons of cargo per day

e. Solution

The numbers of planes used are ...

• 10 C-47s
• 32 C-54s

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Additional comment

For 2 million people that capacity works out to about 0.355 pounds of cargo per person per day.

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