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A plane 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) = 50 + + x 7 34,000 x where x is the ground speed (airspeed ± wind). (a) What is the cost when the ground speed is miles per hour; miles per hour? 470 590 (b) Find the domain of C. (c) Use a graphing calculator to graph the function C = C(x). (d) Create a TABLE with TblStart = 0 and Tbl = 50. (e) To the nearest 50 miles per hour, what ground speed minimizes the cost per passenger?

Sagot :

MrRoyalidn

Tables and graphs can be used to represent a function

  • The cost when ground speed is 470mph and 590mph are $189.48 and $191.91 respectively.
  • The domain of the function is x > 0
  • The ground speed that minimizes the cost is 487.85 mph

Given

[tex]C(x) = 50 + \frac x7 + \frac{34000}{x}[/tex]

A. The cost when ground speed is 470 and 590mph

This means that x = 470 and x = 590

When [tex]x = 470[/tex], we have:

[tex]C(470) = 50 + \frac{470}{7} + \frac{34000}{470}[/tex]

[tex]C(470) = 189.48[/tex]

When [tex]x = 590[/tex], we have:

[tex]C(590) = 50 + \frac{590}{7} + \frac{34000}{590}[/tex]

[tex]C(590) = 191.91[/tex]

B. The domain

[tex]C(x) = 50 + \frac x7 + \frac{34000}{x}[/tex]

For the plane to move, the value of x (i.e. the ground speed) must be greater than 0.

Hence, the domain of the function is x > 0

C. The graph

See attachment for the graph of C(x)

D. The table of values from 0 to 50

So, we have:

[tex]C(0) = 50 + \frac{0}{7} + \frac{34000}{0} = und efin e d[/tex]

[tex]C(10) = 50 + \frac{10}{7} + \frac{34000}{10} =3451.43[/tex]

[tex]C(20) = 50 + \frac{20}{7} + \frac{34000}{20} =1752.85[/tex]

[tex]C(30) = 50 + \frac{30}{7} + \frac{34000}{30} = 1187.62[/tex]

[tex]C(40) = 50 + \frac{40}{7} + \frac{34000}{40} = 905.71[/tex]

[tex]C(50) = 50 + \frac{50}{7} + \frac{34000}{50} = 737.14[/tex]

So, we have:

[tex]\left[\begin{array}{ccccccc}x&0&10&20&30&40&50\\C(x)&&3451.43&1752.85&11.87.62&905.71&737.14\end{array}\right][/tex]

E. The speed that minimizes the cost

We have:

[tex]C(x) = 50 + \frac x7 + \frac{34000}{x}[/tex]

Differentiate the function

[tex]C'(x) = \frac 17 - 34000x^{-2}[/tex]

Equate to 0

[tex]\frac 17 - 34000x^{-2} = 0[/tex]

Collect like terms

[tex]- 34000x^{-2} = -\frac 17[/tex]

[tex]34000x^{-2} = \frac 17[/tex]

Solve for [tex]x^2[/tex]

[tex]x^2 = 34000 \times 7[/tex]

[tex]x^2 = 238000[/tex]

Take square roots

[tex]x = 487.85[/tex]

Hence, the ground speed that minimizes the cost is 487.85 mph

Read more about functions, tables and graphs at:

https://brainly.com/question/13473963

View image MrRoyal
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