Question: Help on a calculus extreme value problem?

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Answer #1:

First, I have to clarify some assumptions.
The coast is straight. The pair of divers are on the coastline presently.
And the course are straight lines up the coast and directly from the point they go in the water, to the wreck.

So you want to minimize time, so write the time equation.
but first total displacement is
Y=.8 and X=.4

First equation to find the launch site (0,k) is
t1=(k-0)/5 (directly walk along the coast k Km)
Second equation is from launch to wreck to find distance
d=sqrt( (.8-k)^2+(.4-0)^2 )
t2=d/3
therefore t=t1+t2
t=k/5+sqrt[ (.8-k)^2 +(.4)^2 ]/3 = k/5 + sqrt[ .64-1.6k +k^2 + .16 ]/3 =
t= k/5 + sqrt[ k^2 -1.6k + .8] = k/5 +sqrt[ (k-.8)^2 ]/3

Minimize over k

Idea is simple, express time walking keeping distance as variable k, and then add time swimming and calculate distance as hypotenuse.
and if V=d/t, t=d/v





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