Q:

# A rock formation is 3 miles due north of its closest point along a straight shoreline. A visitor is staying in a tent that is 4 miles west of that point. The visitor is planning to go from the tent to the rock formation. Suppose the visitor runs at a rate of 6 mph and swims at a rate of 3 mph. How far should the visitor run to minimize the time it takes to reach the rock formation?

Accepted Solution

A:
Answer:   about 2.268 milesStep-by-step explanation:Let x represent the distance the visitor should run. Then the distance he swims will be ...   d = √((4 -x)² +3²) = √(x² -8x +25)The total travel time is given by ...   time = distance / speed   time = x/6 +d/3   time = x/6 +1/3√(x² -8x +25)The time will be minimized at the value of x that makes the derivative zero.   d(time)/dx = 0 = 1/6 +1/6(2x -8)/√(x² -8x +25)   0 = √(x² -8x +25) +2x -8   (8 -2x)² = x² -8x +25 . . . . . . subtract (2x-8), square both sides   3x^2 -24x +39 = 0 . . . . . . subtract the right side   x² -8x +13 = 0 . . . . . . . . . divide by 3   (x -4)² = 3 . . . . . . . . . . . complete the square   x = 4 -√3 ≈ 2.268 . . . . . . . . . . square root, add 4The visitor should run about 2.268 miles to minimize the time._____Additional commentThe ratio of swim rate in water to running rate is 3/6 = 1/2. The angle with respect to straight out from the shoreline will be arcsin(1/2) = 30°, so the distance from the point closest to the rock will be (3 mi)tan(30°) = 1.732 mi. The distance to run is 4 mi - 1.732 mi = 2.268 mi. The angle relation is the generic solution to a problem like this.