A man in a car at location Q on a straight highway is moving with speed v. He decides to reach a point P in a field at a distance d from highway (point M) as shown in the figure. Speed of the car in the field is half to that on the highway. What should be the distance RM, so that the time taken to reach P is minimum?

Let the distance $$RM$$ be $$x$$. And let the distance $$QM$$ be $$L$$. We have, $$PR= \sqrt{x^2+d^2}$$

The total time taken to travel $$PR$$ will be $$\dfrac{\sqrt{x^2+d^2}}{v\div 2} = \dfrac{2\sqrt{x^2+d^2}}{v}$$

Time taken to cover the remaining distance $$L-x$$ from $$Q$$ to $$R$$ will be $$\dfrac{L-x}{v}$$

Thus, the total time,

$$T= \dfrac{L-x}{v} + \dfrac{2\sqrt{x^2+d^2}}{v}$$

$$\Rightarrow T = \dfrac{L-x+2\sqrt{x^2+d^2}}{v}$$

To minimise time, we will take the derivative of $$T$$ with respect to $$x$$, and equate it to $$0$$, we get,

$$\dfrac{dT}{dx} = \dfrac{d}{dx}\left[\dfrac{L-x + 2{(x^2+d^2)}^{1/2}}{v}\right]$$

$$\dfrac{dT}{dx} = 0 - \dfrac{1}{v} + \dfrac{2}{v}\cdot \dfrac{2x}{2}(x^2+d^2)^{-1/2}$$

$$\Rightarrow \dfrac{dT}{dx} = -\dfrac{1}{v}+ \dfrac{2x}{v}\cdot (x^2+d^2)^{-1/2}$$

Equating $$\dfrac{dT}{dx}$$ to zero to get the minimum value, we have,

$$\dfrac{2x}{v}\cdot (x^2+d^2)^{-1/2} = \dfrac{1}{v}$$

$$\dfrac{2x}{\sqrt{x^2+d^2}} = 1$$

$$\Rightarrow {(2x)}^2 = x^2+d^2$$

Which gives $$3x^2 = d^2$$  or  $$x=\dfrac{d}{\sqrt{3}}$$