Page 115 - How To Solve Word Problems In Calculus

Step2
The distance between the ships is determined by the
theorem of Pythagoras:

2
z = x + y 2

Step3
Since ship B is traveling 10 mi/h, after t hours
x = 100 − 10t. Similarly, y = 20t. Thus

2
z = (100 − 10t) + (20t) 2

2
= 10,000 − 2000t + 100t + 400t 2

2
= 500t − 2000t + 10,000
Step4
Because of the radical, it is more convenient to minimize
2
2
z rather than z. The value of t that minimizes z is the same
value that minimizes z.

2
Let F (t) = 500t − 2000t + 10,000
F (t) = 1000t − 2000

0 = 1000t − 2000

t = 2

Intuitively it is obvious that there is some point in time when
the ships are closest. Since t = 2 is the only critical number,
the ships must be closest at 2:00 and the minimum distance
√

2
between them is z = 500(2) − 2000(2) + 10,000 = 8000 =
√
40 5 miles. We can use the second derivative test to confirm
this mathematically.
Step5 (optional)
Since F (t) = 1000 > 0, t = 2 yields a relative minimum.

Since it is the only relative extremum, it is the absolute
minimum.

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