Help with what I think is a differential problem

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SUMMARY

The discussion centers on a differential problem involving a man in a kayak who must determine the optimal point to land on a beach to minimize travel time to his destination. The problem requires the application of calculus, specifically differentiation, to find the minimum total time based on the distances paddled and walked. The total time equation is derived as (1/4)√(x² + 9) + (1/5)(6 - x), where x is the distance from point O to the landing point P. To solve the problem, one must differentiate this equation with respect to x and set the derivative to zero to find the optimal landing distance.

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perjac
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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 [itex]\sqrt{x^2+ 9}[/itex] kilometers. At 4 km/hr, that will require [itex](1/4)\sqrt{x^2+9}[/itex] hours. He then has to walk 6- x kilometers. At 5 km/hr, that will require [itex](1/5)(6- x)[/itex] hours. The total time is [itex](1/4)\sqrt{x^2+ 9}+ (1/5)(6- x)[/itex] hours. You want to minimize that. Take the derivative with respect to x and set it equal to 0.
 

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