Answer :
Let x = the distance between C and D, or 0x1.
We wish to cut down on the amount of time, t, that is required to row to point D and then run the remaining distance to point B.
As a right triangle, ACD. Let s equal the distance between A and D.
s=√(x2+42)
The distance from point D to point B is 1-x.
time equals distance/rate
t = √(x2+42)/6 + (1-x)/8
Take dt/dx, set it to zero, then solve for x to reduce t.
dt/dx is equal to x/(6x2+42) - 1/8.
x/(6√(x2+42)) = 1/8
(6/8)√(x2+42)=x
both sides are square.
(9/16)(x2+42)=x2
(9/16)x2+9 = x2
(7/16)x2=9
x = ±√((9)(16)/7) = ±12/√7 Neither value falls within the range [0, 1].
Verify the endpoints x=0 and x=1.
t(1) = √17/6 - 1/8 = 0.562
t(0) = 4/6+1/8 = 0.792
Therefore, x=1 reduces the time. The ideal course of action is to row directly to point B.
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