Optimization Problem Need Almost Done Stuck

Hi I am stuck on a optimization problem so I am here for some assistance. Heres the question:

Simon Suffers an injury at a campsite on the bank of a canal that is 6√2 km wide. If he is able to paddle his kayak at 5 km/h and pedal a bicycle at 15 km/h along the opposite bank, where should he land on the opposite bank to reach the hospital in minimum time?

here is a diagram that comes with the question:
Attachment 23946

So out of this question I've determined this:
Distance = Velocity * time <- this can be re-arranged to this -> time = Distance / Velocity

distance on kayak: √((6√2)^2 + x^2)
distance on land: (4 - x)

so my formula is this:
Time(x) = (1/15)(4 - x) + (1/5)(√(x^2 + 72))

I differentiated and got this:
Time'(x) = [ (x) / (5 * √(x^2 + 72) ) ] - (1/15)

Now the next step I do is I set my Time'(x) = 0 and solve for x but I feel that I am getting a bad answer. Heres what I have:
t’(x) = 0
(x/(5sqrt(x^2 + 72)) – (1/15) = 0
(x/(5sqrt(x^2 + 72)) = (1/15)
(x)^2 = (1/15)^2(5sqrt(x^2 + 72)^2
x^2 = (1/225)(25)(x^2 + 72)
(x^2) * 225 = ((1/225)(25)(x^2 + 72))*225
(225x^2)/25 = ((5625)(225x^2 + 405000))/25
9x^2 = (225)(9x^2 +16200)
9x^2 = 2025x^2 + 3645000
9x^2 - 2025x^2 = 3645000
-2016x^2 - 3645000
x^2 = -1808.35714x^2

So I cant continue this because as you can see 3645000 does not divide nicely with -2016 and I cant take a square root of a negative number.

If someone could point out where the mistake is that would be GREATLY !! appreciated!!!

Thanks!!!

The calculation can be made much more simpler rather than finding a mistake in your approach to the problem.
x/5*sqrt(x^2+72)=1/15
15x=5*sqrt(x^2+72)
3x=sqrt(x^2+72)
9*x^2=x^2+72
8*x^2=72
x^2=9
x=3

Ok thanks shyamnair3. I should of simplified more instead of over complicating it.

Now the 3 I can plug into my first equation to get the minimal time to answer this question right?

Can someone please explain how the derivative of this is taken? I have been stuck on this exact question for a very very long time...

my equation is t=(3(sqrt(x^2+72)+4-x)/15

I need to know how to get the derivative of this. I have looked up this exact question is a few places and nobody actually shows how to derive this. Everybody ends up with an answer of x=3 and I have tried an exhausting amount of methods to try and get that.

anybody??

$\displaystyle t=\frac{3(\sqrt{x^2+72}+4-x)}{15}=\frac{\sqrt{x^2+72}+4-x}{5}$

$\displaystyle t=\frac{\sqrt{x^2+72}+4-x}{5}$

I'll just show how to get the derivative of $\displaystyle \sqrt{x^2+72}$ the rest is straightforward
$\displaystyle y=\sqrt{x^2+72}$

$\displaystyle \text{Let } u=x^2+72$

$\displaystyle \frac{du}{dx}=2x$

$\displaystyle y=\sqrt{u}$

$\displaystyle \frac{dy}{du}=0.5u^{-0.5}$

$\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$

$\displaystyle \frac{dy}{dx}=0.5u^{-0.5}\cdot 2x$

$\displaystyle \frac{dy}{dx}=0.5(x^2+72)^{-0.5}\cdot 2x$

Thanks a lot man, I can see where I messed up now! Appreciate it!!