Problem 69

A man on an island 12 miles south of a straight beach wishes to reach a point on shore 20 miles east. If a motorboat, making 20 miles per hour, can be hired at the rate of $2.00 per hour for the time it is actually used, and the cost of land transportation is $0.06 per mile, how much must he pay for the trip?

Solution:
Distance traveled by boat:
s = \sqrt{12^2 + x^2}
s = \sqrt{144 + x^2}

Note: time = distance/speed

Total cost of travel:
C = \dfrac{s}{20}(2) + (20 – x)0.06

C = \frac{1}{10}\sqrt{144 + x^2} + 1.2 – 0.06x

\dfrac{dC}{dx} = \dfrac{1}{10} \left( \dfrac{2x}{2\sqrt{144 + x^2}} \right) - 0.06 = 0

\dfrac{x}{10\sqrt{144 + x^2}} = 0.06

x = 0.6\sqrt{144 + x^2}
x^2 = 0.36(144 + x^2)
0.64x^2 = 51.84
x = 9 \text{ miles}

C = \frac{1}{10}\sqrt{144 + 9^2} + 1.2 – 0.06(9)
C = \$ 2.16 \,\, answer

Problem 70

A man in a motorboat at A (Figure 42) receives a message at noon calling him to B. A bus making 40 miles per hour leaves C, bound for B, at 1:00 PM. If AC = 40 miles, what must be the speed of the boat to enable the man to catch the bus.

Solution:
distance = speed × time
(rt)^2 = 40^2 + 40^2(t – 1)^2
r^2t^2 = 1600 + 1600t^2 – 3200t + 1600

r^2 = \dfrac{1600(t^2 - 2t + 2}{t^2}

r = \dfrac{40}{t}\sqrt{t^2 - 2t + 2}

\dfrac{dr}{dt} = \dfrac{40}{t}\dfrac{2t - 2}{2\sqrt{t^2 - 2t + 2}} + \dfrac{-40}{t^2}\sqrt{t^2 - 2t + 2} = 0

\dfrac{t - 1}{\sqrt{t^2 - 2t + 2}} = \dfrac{1}{t}\sqrt{t^2 - 2t + 2}

t^2 – t = t^2 – 2t + 2
t = 2 \text{ hours}

r = \dfrac{40}{2}\sqrt{2^2 - 2(2) + 2}
r = 20 \sqrt{2} \text{ miles/hour}
r = 28.28 \text{ miles/hour } \,\,            answer

Problem 71

In Problem 70, if the speed of the boat is 30 miles per hour, what is the greatest distance offshore from which the bus can be caught?

Solution:
By Pythagorean Theorem:
y = \sqrt{900t^2 − 1600(t − 1)^2}

\dfrac{dy}{dt} = \dfrac{1800t - 3200(t - 1)}{2\sqrt{900t^2 − 1600(t − 1)^2}} = 0

9t – 16(t – 1) = 0
7t = 16
t = 16/7

y = \sqrt{900(16/7)^2 − 1600(16/7 − 1)^2}
y = 45.35 \text{ miles } \,\,            answer