See the diagram please.
The car travels with a speed v on the field and with a speed v * n on the main highway. The car starts from A and travels on the road upto C and then travels along CB on the field till B.
Let the distance AD = d. The perpendicular distance of point B on the field from the highway is DB = L. Let the distance from point D of C (turning point) be x.
time taken for traveling distance AC = t1 = (d - x) / (v n )
time duration for travelling distance CB = t2 = √(L²+x²) / v
Total time duration = T = t1 + t2 = √(L² + x²) / v + (d - x) / (v n)
We want to minimize T wrt x, so differentiate and dT /dx = 0
dT/dx = 2 x / [2 v √(L² + x²) ] - 1 / (v n)
equating it to 0, we get √(L²+x²) = n x
=> x = L / (n² - 1) This is the answer.
Now we can confirm that T is minimum at this value of x by finding the second derivative of T wrt x.
d²T / dx² = [L² - x²] / [v (L² +x²)^1.5] > 0
as L > x as n > 1
Hence we get that at the distance x = L /(n² -1) from the point D, the car must turn in to the field, in order to reach there in the least amount of time.