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

According to the Pythagorean theorem, for a right - triangle with sides \(a=x - 3\), \(b = x\) and hypotenuse \(c = 15\), we have \((x-3)^{2}+x^{2}=15^{2}\).
Expand \((x - 3)^{2}\): \(x^{2}-6x + 9+x^{2}=225\).
Combine like terms: \(2x^{2}-6x+9 - 225=0\), so \(2x^{2}-6x - 216 = 0\).
Divide by 2: \(x^{2}-3x - 108=0\).
Factor the quadratic equation: \(x^{2}-3x - 108=(x - 12)(x + 9)=0\).
Solve for \(x\): \(x=12\) or \(x=-9\). Since distance cannot be negative, \(x = 12\).

Step2: Calculate the total distance of the alternative route

The eastern leg is \(x-3=12 - 3 = 9\) miles and the northern leg is \(x = 12\) miles.
The total distance of the alternative route is \(9 + 12=21\) miles.

Step3: Calculate the time for each route

For the mountain route:
The formula for time \(t=\frac{d}{v}\), where \(d = 15\) miles and \(v = 45\) mph. So \(t_{1}=\frac{15}{45}=\frac{1}{3}\) hours.
For the alternative route:
The total distance \(d = 21\) miles and \(v = 70\) mph. So \(t_{2}=\frac{21}{70}=\frac{3}{10}\) hours.
Compare \(\frac{1}{3}\approx0.333\) and \(\frac{3}{10}=0.3\)

Answer:

(a) The total distance of the alternative route is \(21\) miles.
(b) Since \(\frac{3}{10}<\frac{1}{3}\), Justin should take the alternative (east - then - north) route.