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A man commutes to work a distance of 4040 miles and returns on the same route at the end of the day. His average rate on the return trip is 2525 miles per hour faster than his average rate on the outgoing trip. Write the total​ time, T, in​ hours, devoted to his outgoing and return trips as a function of his rate on the outgoing​ trip, x. Then find and interpret Upper P (30 ).T30).

Respuesta :

Answer:

[tex]T(x) = \frac{80x+1000}{x^2+25x}[/tex]

[tex]T(30) = \frac{3400}{1650} = 2.061[/tex]

Step-by-step explanation:

If the rate is x, then the rate of the return is x+25. In both trips, he has 40 miles to travel, therefore, it will take 40/x hours in the first trip and 40/x+25 for the return. As a consecuence

[tex]T(x) = \frac{40}{x} + \frac{40}{x+25} = \frac{40 ( x+(x+25))}{x(x+25)} = \frac{80x+1000}{x^2+25x}[/tex]

Also,

[tex]T(30) = \frac{80*30+1000}{30^2+25*30} = \frac{3400}{1650} = 2.061[/tex]

This means that if he starts at 30 miles per hour, it will take a bit more than 1 hour to be there: 1 hour and 20 minutes in the first trip, and a bit more than 40 minutes (40/55 hours) in the return.

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