A man launches his boat from point A on a bank of a straight river A man launches his boat from point A on a bank of a straight river, 3 km wide, and wants to reach point B, 3 km downstream on the opposite bank, as quickly as possible. He could row his boat directly across the river to point C and then run to B, or he could row directly to B, or he could row to some point D between C and B and then run to B. If he can row 6 km/h and run 8 km/h, where should he land to reach B as soon as possible? (We assume that the speed of the water is negligible compared to the speed at which the man rows.) 
Attempt:
 time= distance/rate
t(x)= t(row)+t(run)
= $\frac{1}{6}\sqrt{9+x^2} + \frac{1}{8}(3-x)$
$T'(x)= \frac{x}{6\sqrt{9+x^2}} - \frac{1}{8}$
$T'(x)=0$ when $x/(6\sqrt{9+x^2})=1/8$
$x=9\sqrt{7}
and when I sub in T'(0)=-1/8 and T'(8)=0.3105 >0 so min at x= 9 \sqrt{7} but my homework websites said it is wrong and that they want an exact value. Can someone give me a hand.
 A: The problem is that the critical point you found is not on the domain! The values for any solution must be $x\in[0,3]$. Since your critical point lies outside the domain, you have to check the endpoints for your solution. In this case the answer should be $x=3$. In other words, he should row all the way!
A: Let x = the distance from C to D, 0≤x≤1
We want to minimize the time, t, it takes t row to point D then run the rest of the distance to point B
ACD form a right triangle. Let s = the distance from A to D,
s=√(x2+42)
1-x is the distance from point D to B
time = distance/rate
t = √(x2+42)/6 + (1-x)/8
To minimize t, take dt/dx, set that to zero and solve for x
dt/dx = x/(6√(x2+42)) - 1/8
x/(6√(x2+42)) = 1/8
(6/8)√(x2+42)=x
Square both sides
(9/16)(x2+42)=x2
(9/16)x2+9 = x2
(7/16)x2=9
x = ±√((9)(16)/7) = ±12/√7 Neither value is in the interval [0,1]
Check endpoints, x=0 and x=1
t(1) = √17/6 - 1/8 = 0.562
t(0) = 4/6+1/8 = 0.792
So x=1 minimizes the time. The best solution is to row straight to point B
