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Sagot :
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

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