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From point 'A' located on a highway, one has to get by a car as soon as possible to point 'B' located in the field at a distance ' l ' , from point 'D' .

if the car moves n times slower in the field at what distance x from D one must turn off the highway.



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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.

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