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(the following answer is included essentially in "The Feynman LECTURES ON PHYSICS-Mechanics, Radiation & Heat ,Vol. 1, 26-3 Fermat's principle of least time.)
Suppose you are at point A in the land and a screaming girl is at point B in the sea. You can run with a speed $\:v_{1}\:$ on the land greater than the speed $\:v_{2}\:$ you can swim in the sea. At a moment you decide to follow the path ACB spending time $\: t_{1}=AC/v_{1}\:$ running on the land and time $\: t_{2}=CB/ v_{2}\:$ swimming in the sea, that is total time
\begin{equation}t_{tot}=t_{1}+t_{2}=\dfrac{AC}{v_{1}}+\dfrac{CB}{v_{2}}\tag{01}\end{equation}
But after a while you change your mind and decide to displace the point C on shoreline a little to the right to the point D. But then you are wondering if by such a displacement you shorten the total time or not.
For infinitesimally small displacement $\:CD\equiv \Delta x\:$ you can do the following approximations :
\begin{equation}AE \approx AC \qquad \theta_{1}^{\prime} \approx \theta_{1} \qquad BZ \approx BD \qquad \theta_{2}^{\prime} \approx \theta_{2}\tag{02}\end{equation}
You realize that on one hand you decrease the swimming distance by $$BC-BD \approx CZ=\Delta x \cdot \sin\theta_{2}\tag{03}$$so changing (decreasing) the swimming time by $$\Delta t_{2}=t_{2}^{\prime}-t_{2}=-\Delta x \cdot \dfrac{\sin\theta_{2}}{v_{2}}\tag{04}$$On the other hand you increase the running distance by $$AD-AC \approx DE=\Delta x \cdot \sin\theta_{1}\tag{05}$$so changing (increasing) the running time by $$\Delta t_{1}=t_{1}^{\prime}-t_{1}=+\Delta x \cdot \dfrac{\sin\theta_{1}}{v_{1}}\tag{06}$$
So, balancing, the total time change is$$\Delta t_{tot}=t_{tot}^{\prime}-t_{tot}=\left(t_{2}^{\prime}+t_{1}^{\prime} \right)-\left(t_{2}+t_{1}\right)=\Delta t_{2}+\Delta t_{1}= \Delta x \cdot\left(\dfrac{\sin\theta_{1}}{v_{1}}- \dfrac{\sin\theta_{2}}{v_{2}}\right)\tag{07}$$This means that if $$\left(\dfrac{\sin\theta_{1}}{v_{1}}- \dfrac{\sin\theta_{2}}{v_{2}}\right) > 0\tag{08}$$then moving infinitesimally to the right, $\: \Delta x >0 \:$, we increase the time while moving to the left, $\: \Delta x <0 \:$, we decrease the time. So in case that the condition (08) is valid then in order to find a shorter time we must search to the left of point C.
If $$\left(\dfrac{\sin\theta_{1}}{v_{1}}- \dfrac{\sin\theta_{2}}{v_{2}}\right) < 0\tag{09}$$then moving infinitesimally to the right, $\: \Delta x >0 \:$, we decrease the time while moving to the left, $\: \Delta x <0 \:$, we increase the time. So in case that the condition (09) is valid then in order to find a shorter time we must search to the right of point C.
But finally, if$$\left(\dfrac{\sin\theta_{1}}{v_{1}}- \dfrac{\sin\theta_{2}}{v_{2}}\right) = 0\tag{10}$$
then either moving to the right,$\: \Delta x >0 \:$, or moving to the left , $\: \Delta x <0 \:$, the change is infinitesimally zero. This is the definition of the extreme points of a function.So, condition (10) is the one of the shortest time and if you are a light ray then in terms of refraction indices
$$v_{1}=c_{1}=\dfrac{c_{0}}{n_{1}}, \quad v_{2}=c_{2}=\dfrac{c_{0}}{n_{2}}\tag{11}$$and (10) is Snell's Law
$$\bbox[#FFFF88,12px]{n_{1}\sin\theta_{1}=n_{2}\sin\theta_{2} \qquad \textbf{(Snell's Law)}} \tag{12}$$
