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Chapter 3: Problem 111

Cost of Transatlantic Travel A Boeing 747 crosses the Atlantic Ocean (3000 miles) with an airspeed of 500 miles per hour. The cost \(C\) (in dollars) per passenger is given by $$ C(x)=100+\frac{x}{10}+\frac{36,000}{x} $$ where \(x\) is the ground speed (airspeed \(\pm\) wind). (a) What is the cost per passenger for quiescent (no wind) conditions? (b) What is the cost per passenger with a head wind of 50 miles per hour? (c) What is the cost per passenger with a tail wind of 100 miles per hour? (d) What is the cost per passenger with a head wind of 100 miles per hour?

Short Answer

Expert verified
The cost per passenger is \(222 (no wind), \)225 (head wind 50 mph), \(220 (tail wind 100 mph), and \)230 (head wind 100 mph).

Step by step solution

01

Understand the Given Function

The cost per passenger is given by the formula: \[ C(x) = 100 + \frac{x}{10} + \frac{36,000}{x} \] where \( x \) is the ground speed.
02

Calculate the Cost for Quiescent Conditions

For no wind conditions, the ground speed is equal to the airspeed. Given that the airspeed is 500 miles per hour, we set \( x = 500 \). Substitute \( x = 500 \) into the cost function: \[ C(500) = 100 + \frac{500}{10} + \frac{36,000}{500} \]Simplify it: \[ C(500) = 100 + 50 + 72 = 222 \]So, the cost per passenger for no wind conditions is \$222.
03

Calculate the Cost with a Head Wind of 50 mph

For a head wind of 50 miles per hour, the ground speed will be reduced by 50 miles per hour. Therefore, the ground speed is \( x = 500 - 50 = 450 \) miles per hour. Substitute \( x = 450 \) into the cost function: \[ C(450) = 100 + \frac{450}{10} + \frac{36,000}{450} \]Simplify it: \[ C(450) = 100 + 45 + 80 = 225 \]So, the cost per passenger with a head wind of 50 miles per hour is \$225.
04

Calculate the Cost with a Tail Wind of 100 mph

For a tail wind of 100 miles per hour, the ground speed will be increased by 100 miles per hour. Therefore, the ground speed is \( x = 500 + 100 = 600 \). Substitute \( x = 600 \) into the cost function: \[ C(600) = 100 + \frac{600}{10} + \frac{36,000}{600} \]Simplify it: \[ C(600) = 100 + 60 + 60 = 220 \]So, the cost per passenger with a tail wind of 100 miles per hour is \$220.
05

Calculate the Cost with a Head Wind of 100 mph

For a head wind of 100 miles per hour, the ground speed will be reduced by 100 miles per hour. Therefore, the ground speed is \( x = 500 - 100 = 400 \). Substitute \( x = 400 \) into the cost function: \[ C(400) = 100 + \frac{400}{10} + \frac{36,000}{400} \]Simplify it: \[ C(400) = 100 + 40 + 90 = 230 \]So, the cost per passenger with a head wind of 100 miles per hour is \$230.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cost Function
If you replace `x` with the actual ground speed, you can find the passenger cost based on different wind conditions. Notice how each component term: \(100 + \frac{x}{10} + \frac{36,000}{x}\), affects the cost distinctively.
Ground Speed
Ground speed is the actual speed at which the aircraft moves over the ground. It is different from airspeed because wind plays a crucial role. For instance, if the airspeed is 500 mph and there is no wind, then the ground speed is also 500 mph. However, headwinds and tailwinds will affect this speed.
  • With a tailwind, the ground speed increases because the wind pushes the aircraft forward.
  • With a headwind, the ground speed decreases because the wind hampers the aircraft's progress.
Understanding how to calculate ground speed in different wind conditions allows you to use the correct value of `x` in the cost function.
Wind Effect on Travel Cost
Wind has a significant effect on travel cost. It influences the ground speed, which in turn affects the cost per passenger. Here鈥檚 how different wind conditions can alter the costs:
  • In quiescent (no wind) conditions, ground speed equals airspeed.
  • With headwind, ground speed decreases, leading to higher travel times and possibly higher costs.
  • With tailwind, ground speed increases, leading to shorter travel times and possibly decreased costs.
For instance, a headwind of 50 mph reduces the ground speed from 500 to 450 mph and changes the cost calculation.
Algebraic Manipulation
Algebraic manipulation involves substituting different values into the cost function and simplifying the result to understand varying costs under different conditions. Here鈥檚 how to do it: 1. **Substitution:** Replace `x` in the cost function with the actual ground speed. 2. **Simplification:** Simplify each term separately, then add the simplified terms. For example, for quiescent conditions with a ground speed of 500 mph: \[ C(500) = 100 + \frac{500}{10} + \frac{36,000}{500} \] Simplify each term: \[ C(500) = 100 + 50 + 72 = 222 \] This process allows you to see how the cost changes based on the ground speed under various wind conditions.

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