From point A located on a highway (figure) one has to get by car as soon as possible to point B located in the field at a distance `l` from the highway. It is known that the car moves in the field `eta` times slower than on the highway. At what distance from point D one must turn off the highway?

Explanation

Suppose at x distance from D the car turns off the highway.
Let v be the speed of car on highway, then its speed on field will be `v//eta`.
Time of motion = Time to travel on highway+Time to travel on field
Time to travel on highway, `T_(highway)=(AD-x)/(v)`
Time to travel on field, `T_(field)=(sqrt(x^2+l^2))/(v//eta)`
Here, total time, `T=((AD-x)/(v))+(sqrt(x^2+l^2))/(v//eta)` ...(i)
If T to be minimum, `(dT)/(dx)=0`
`(dt)/(dx)=(d)/(dx)[(AD-x)/(v)+(eta)/(v)(x^2+l^2)^(1//2)]` [AD is constant]
`0=-1/v+eta/vxx1/2(x^2+l^2)^(-1//2)xx2x`
`eta^2x^2=x^2+l^2impliesx=(1)/(sqrt(eta^2-1))`
Here we need not to do double derivative test as we know the maximum time for this case is not defined.