8. You are out for a day of sailing and plan on coming back to shore for lunch. The boat is currently anchored 2 miles from shore and 6 miles down shore from the 2 mi restaurant. a. If you can sail at a speed of 4mph and walk along the shore at a speed of 3 mph, at what point along the shore should you dock and start walking in order to reach the restaurant in the minimum time? How long will it take you to reach the restaurant? b. If you are able to walk at the same speed, but the wind has picked up and you can sail at a faster speed compared to the problem above, would the optimal place to start walking be closer to or farther from the restaurant compared to your answer on part a? Explain your answer and reasoning using complete sentences and correct grammar.
8. You are out for a day of sailing and plan on coming back to shore for lunch. The boat is currently anchored 2 miles from shore and 6 miles down shore from the 2 mi restaurant. a. If you can sail at a speed of 4mph and walk along the shore at a speed of 3 mph, at what point along the shore should you dock and start walking in order to reach the restaurant in the minimum time? How long will it take you to reach the restaurant? b. If you are able to walk at the same speed, but the wind has picked up and you can sail at a faster speed compared to the problem above, would the optimal place to start walking be closer to or farther from the restaurant compared to your answer on part a? Explain your answer and reasoning using complete sentences and correct grammar.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Given that a boat is anchored 2 miles from shore and 6 miles from the restaurant.
VIEWLet x be the point along the shore to dock and start walking to the restaurant.
VIEWThe speed of the boat is 4mph. Therefore, the time required to sail is
VIEWThe total time T(x) will be,
VIEWNote: We use the second derivative test to get the minimum value,
VIEWPart b:
VIEWAnswer:
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