Help with what I think is a differential problem

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The discussion revolves around solving a differential problem involving a man in a kayak needing to reach a destination on a beach. The scenario includes calculating the optimal point on the beach to land, minimizing travel time by considering both paddling and walking speeds. The total time is expressed as a function of the distance from the nearest point on the beach, incorporating the Pythagorean theorem for distance calculations. To find the minimum time, the derivative of the total time function must be taken and set to zero. This approach provides the necessary steps to solve the problem effectively.
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Hi All,

Did my maths B exam the other day but was absolutely stumped by one question. Spent half an hour thinking about it only to give up.

Question:

A man on a kayak (K) is 3 kilometres out to sea from the nearest point, (O) of a straight beach. His destination (D) is 6 kilometres along the beach from O. The fastest he can paddle is 4 km/hr and his maximum walking speed is 5km/h. How far from O should he go ashore to reach his destination in the least possible time?

Anyways the question was alongside a whole heap of differential problems so I assume you would need to differentiate an equation, find the min SP and that should tell you the distance but I have no idea how to get to the equation or if I was even on the right track.

This was the only question I couldn't answer and it is driving me nuts.

Thanks,

Perjac
 
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Let P be the point at which the kayak lands on the beach. Let x be the distance from O to P. By the Pythagorean theorem, he must have paddled a distance \sqrt{x^2+ 9} kilometers. At 4 km/hr, that will require (1/4)\sqrt{x^2+9} hours. He then has to walk 6- x kilometers. At 5 km/hr, that will require (1/5)(6- x) hours. The total time is (1/4)\sqrt{x^2+ 9}+ (1/5)(6- x) hours. You want to minimize that. Take the derivative with respect to x and set it equal to 0.
 
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