Question

justin must travel from summersville to clayton. he can drive 15 mi through the mountains at 45 mph. or he can drive east and then north on superhighways at 70 mph. the alternative route forms a right angle as shown in the diagram. the eastern leg is 3 mi less than the northern leg. (a) find the total distance justin would travel in going the alternative route. (b) if justin wants to minimize the time of the trip, which route should he take?

Explanation:

Step1: Use Pythagorean theorem

Let the northern leg be \(x\) miles. Then the eastern leg is \((x - 3)\) miles. By the Pythagorean theorem \(15^{2}=(x - 3)^{2}+x^{2}\).
Expand: \(225=x^{2}-6x + 9+x^{2}\).
Combine like terms: \(2x^{2}-6x-216 = 0\).
Divide by 2: \(x^{2}-3x - 108=0\).
Factor: \((x - 12)(x + 9)=0\).
Since distance can't be negative, \(x = 12\).
The eastern leg is \(x-3=12 - 3=9\) miles.

Step2: Calculate total distance of alternative route

The total distance of the alternative route is \(9 + 12=21\) miles.

Step3: Calculate time for mountain route

Time \(t_{1}=\frac{d}{v}\), where \(d = 15\) miles and \(v = 45\) mph. So \(t_{1}=\frac{15}{45}=\frac{1}{3}\) hours.

Step4: Calculate time for alternative route

Time \(t_{2}=\frac{d}{v}\), where \(d = 21\) miles and \(v = 70\) mph. So \(t_{2}=\frac{21}{70}=0.3\) hours.

Answer:

(a) \(21\) miles.
(b) He should take the alternative route (east - then - north) since \(t_{1}=\frac{1}{3}\approx0.33\) hours and \(t_{2}=0.3\) hours, and \(0.3<0.33\).