The starting point is 60m horizontally from the end point of the climb.
The first part of the route is 160m from the starting point but contains flat rocks, and the climber can't climb very fast down (40m/h).
The second part of the course (65m) is a much better climbing surface and the climber can climb down at a rate of (122m/h).
Help the climber to find the fastest way down the cliff.
Without factoring in the differences in speed you would just use Pythagoras to get $$d(x) = \sqrt{(160m+65m)^2 + 60m^2}$$
However my best attempt to try and form an equation to derive and find the critical points is $$d(x) = \sqrt{60^2 + \left(\frac{160x}{40} + \frac{65x}{122}\right)^2}$$
Which doesn't seem right. Any help would be much appreciated.
$$T = t_1 + t_2 = \frac{\sqrt{160^2 + h^2}}{40} + \frac{\sqrt{65^2 + (60-h)^2}}{112}$$
Take derivative of $T$ with respect to $h$, set to zero to find optimizing $h$.
Answer: $h = 26.52 m,$ measured from the top.
(If you're clever, you could use Snell's Law!)
Your error was in just computing the distance. Instead, you are to minimize time, and the total time consists of two portions. The sum of these two times must be minimized.