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let s(1) = the distance the horse travel before it was watered. s(2) = the distance the horse travel after it was watered. since t = s / v

so total time, T = t(1) + 15min + t(2) T = [s(1)*60]/ 10 + [s(2)*60]/ 12 + 15 min = 6s(1) + 5s(2) + 15 (all in minutes)

the shortest T taken is when s(1) is minimum and hence, it goest straight to the north s(1) = 6km then s(2) become square root of (8^{2}+10^{2}) = 12.806km

hence T = 6(6) + 5(12.806) + 15 T = 115.03 minutes. ?

w = the distance West of the cabin on the river where the horse waters. Dt = distance traveled while thirsty [at 10km/hr, or 6 min/km] Dw = distance traveled while watered [at 12 km/hr, or 5 min/km]

Return time is

[1] Tr = [6 Dt + 15 + 5 Dw] minutes

Minimize Tr with respect to w.

[2] Dt = [(8-w)^{2} + 6^{2}]^{.5} - on the way to the river [3] Dw = [w^{2} + 10^{2}]^{.5} - from the river back to the cabin

It's clear that going straight north to the river [w=8] and going straight south from the river to the cabin [w=0] take longer than needed. So somewhere between 8 and 0 w has a value w_{min} that minimizes Tr.

The roughest approximation for w_{min} is 4 km [halfway], but that's too low because PT starts north of the house If the speeds to and from the river were the same, w_{min} would simply be 5 km - the route of shortest distance. But since it's [slightly] better to travel a little farther watered, w_{min} will be somewhat greater than [west of] 5.

Here's what happens

plug [2] and [3] into [1] and differentiate with respect to w.
Use the chain rule for differentiating: if f(x) = f[g(x)], first set u=g(x); then df/dx = df/du x dy/dx.

Set dTr/dw = 0

Solve for w, getting w_{min} = 5.40889... km west of cabin

evaluate [1]-[3] to get Tr(w_{min}) = 111.0589 ... minutes

Re Chuck Norris,

Spoiler for You can lead a horse to drink

But even Chuck can't make him water.

0

The greatest challenge to any thinker is stating the problem in a way that will allow a solution. - Bertrand Russell

let s(1) = the distance the horse travel before it was watered. s(2) = the distance the horse travel after it was watered. since t = s / v

so total time, T = t(1) + 15min + t(2) T = [s(1)*60]/ 10 + [s(2)*60]/ 12 + 15 min = 6s(1) + 5s(2) + 15 (all in minutes)

the shortest T taken is when s(1) is minimum and hence, it goest straight to the north s(1) = 6km then s(2) become square root of (8^{2}+10^{2}) = 12.806km

hence T = 6(6) + 5(12.806) + 15 T = 115.03 minutes. ?

Your error is assuming when s(1) is minimum, then T will be also minimum. If you calculate in a math way, he shouldn't go straight to north, but 2.82 km. east of it.

As I 've forgotten the derivation of sqrt's, I reached result of 111 minutes by trying 2,3,4 for 2.82 value, and giving 3 km. to east travel, I got 111. Thanks to Bonanova, as he reminded me derivations, and I managed to get the result without try and error.

Really, this is a math problem and it isn't possible to get it without derivations.

I say... Hasn't Professor any water at home? Per Bonanova's calculations, the distance they traveled from their stop to water is only slightly shorter than the distance from that rest point to home. (It's something like 7 km vs. 9 km.) And if the horse refuses to go in any other direction, but towards the water, wouldn't it also insist on taking the shortest path to the river?

It started to heavily down pour and the Professor used his shirt to collect water to quench the horses thirst, then they returned home in a striaght line.

It started to heavily down pour and the Professor used his shirt to collect water to quench the horses thirst, then they returned home in a striaght line.