Climber starts at the top of cliff, finding fastest way to get down. (Optimization calculus) > 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. Optimization Sketch to better explain the problem

$$T = t_1 + t_2 = \frac{\sqrt{160^2 + h^2}}{40} + \frac{\sqrt{65^2 + (60-h)^2}}{112}$$

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!)

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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.