Question

a man commutes to work a distance of 40 miles and returns on the same route at the end of the day. his average rate on the return trip is 25 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 t(35). hint: time traveled = \\(\frac{distance\\ traveled}{rate\\ of\\ travel}\\).
t(x) = \\(\square\\) (simplify your answer.)

Explanation:

Step1: Define Rates and Distances

Outgoing rate: \( x \) mph, distance: 40 miles. Return rate: \( x - 25 \) mph (wait, no—wait, the problem says "his average rate on the return trip is 25 miles per hour faster than his average rate on the outgoing trip"? Wait, no, re-reading: "His average rate on the return trip is 25 miles per hour faster than his average rate on the outgoing trip." Wait, no, original: "His average rate on the return trip is 25 miles per hour faster than his average rate on the outgoing trip"? Wait, no, the user's text: "His average rate on the return trip is 25 miles per hour faster than his average rate on the outgoing trip." Wait, no, let's parse: "A man commutes to work a distance of 40 miles and returns on the same route... His average rate on the return trip is 25 miles per hour faster than his average rate on the outgoing trip, \( x \)." So outgoing rate: \( x \) mph, return rate: \( x + 25 \) mph? Wait, no—wait, "faster than his average rate on the outgoing trip"—so return rate is \( x + 25 \)? Wait, no, the problem says: "his average rate on the return trip is 25 miles per hour faster than his average rate on the outgoing trip, \( x \)." So outgoing rate: \( x \) mph, return rate: \( x + 25 \) mph? Wait, no, wait the distance is 40 miles each way. Time for outgoing trip: \( \frac{40}{x} \) hours. Time for return trip: \( \frac{40}{x + 25} \) hours? Wait, no, wait the problem says: "his average rate on the return trip is 25 miles per hour faster than his average rate on the outgoing trip"—so if outgoing is \( x \), return is \( x + 25 \). Then total time \( T(x) \) is outgoing time plus return time: \( T(x) = \frac{40}{x} + \frac{40}{x + 25} \). Wait, but let's check the hint: Time = Distance / Rate. So outgoing distance 40, rate \( x \), so time \( \frac{40}{x} \). Return distance 40, rate is \( x + 25 \) (since return is 25 mph faster than outgoing). So total time \( T(x) = \frac{40}{x} + \frac{40}{x + 25} \). Now, simplify this: find a common denominator, which is \( x(x + 25) \). So \( T(x) = \frac{40(x + 25) + 40x}{x(x + 25)} = \frac{40x + 1000 + 40x}{x^2 + 25x} = \frac{80x + 1000}{x^2 + 25x} \). We can factor numerator: \( 40(2x + 25) \), denominator: \( x(x + 25) \). Wait, but maybe the problem was "return trip is 25 mph slower"? Wait, no, the text says "faster". Wait, maybe I misread. Let's re-express the problem: "A man commutes to work a distance of 40 miles and returns on the same route at the end of the day. His average rate on the return trip is 25 miles per hour faster than his average rate on the outgoing trip, \( x \). 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 \( T(35) \)."

So Step1: Determine time for each trip. Outgoing: distance 40, rate \( x \), so time \( \frac{40}{x} \). Return: distance 40, rate \( x + 25 \) (since return is 25 mph faster), so time \( \frac{40}{x + 25} \).

Step2: Total time \( T(x) = \frac{40}{x} + \frac{40}{x + 25} \). Simplify by combining fractions:

\( T(x) = \frac{40(x + 25) + 40x}{x(x + 25)} = \frac{40x + 1000 + 40x}{x^2 + 25x} = \frac{80x + 1000}{x^2 + 25x} \). We can factor numerator and denominator: numerator \( 40(2x + 25) \), denominator \( x(x + 25) \). Alternatively, factor 80x + 1000: 40(2x + 25), denominator x(x + 25). So \( T(x) = \frac{40(2x + 25)}{x(x + 25)} \) or \( \frac{80x + 1000}{x^2 + 25x} \).

Now, find \( T(35) \): substitute \( x = 35 \) into \( T(x) \).

First, compute outgoing time: \( \frac{40}{35} = \frac{8}{7} \) hours (≈1.1429 hours). Return rate: \( 35 + 25 = 60 \) mph. Return time: \( \frac{40}{60} = \frac{2}{3} \) hours (≈0.6667 hours). Total time: \( \frac{8}{7} + \frac{2}{3} = \frac{24 + 14}{21} = \frac{38}{21} ≈1.8095 \) hours. Alternatively, using the simplified formula: \( T(35) = \frac{80(35) + 1000}{(35)^2 + 25(35)} = \frac{2800 + 1000}{1225 + 875} = \frac{3800}{2100} = \frac{38}{21} ≈1.8095 \) hours.

Interpretation: When the outgoing rate is 35 mph, the total commute time (outgoing and return) is \( \frac{38}{21} \) hours (or approximately 1.81 hours).

But first, let's confirm the function \( T(x) \). So the function is \( T(x) = \frac{40}{x} + \frac{40}{x + 25} \), which simplifies to \( \frac{80x + 1000}{x^2 + 25x} \) or \( \frac{40(2x + 25)}{x(x + 25)} \).

So Step1: Calculate outgoing time: \( \frac{40}{x} \).

Step2: Calculate return time: \( \frac{40}{x + 25} \) (since return rate is \( x + 25 \)).

Step3: Total time \( T(x) = \frac{40}{x} + \frac{40}{x + 25} \).

Simplify:

\( T(x) = \frac{40(x + 25) + 40x}{x(x + 25)} = \frac{40x + 1000 + 40x}{x^2 + 25x} = \frac{80x + 1000}{x^2 + 25x} \).

Now, for \( T(35) \):

Substitute \( x = 35 \) into \( T(x) \):

\( T(35) = \frac{80(35) + 1000}{(35)^2 + 25(35)} = \frac{2800 + 1000}{1225 + 875} = \frac{3800}{2100} = \frac{38}{21} ≈1.8095 \) hours.

Interpretation: When the man's outgoing speed is 35 mph, his return speed is 60 mph (35 + 25), and the total time for the round trip (commuting to work and back) is \( \frac{38}{21} \) hours, approximately 1.81 hours.

Answer:

First, the function \( T(x) = \frac{80x + 1000}{x^2 + 25x} \) (or \( \frac{40}{x} + \frac{40}{x + 25} \)). Then \( T(35) = \frac{38}{21} \) (or approximately 1.81) hours. The interpretation is that with an outgoing rate of 35 mph (return rate 60 mph), the total commute time is \( \frac{38}{21} \) hours.