Question

A forest ranger is in a forest 1 miles north of a straight road. A car is located on the road at a point 5 miles east of the ranger. If the forest ranger can walk 2 miles per hour in the forest and 4 miles per hour along the road, toward what point on the road should the ranger walk in order to minimize the time needed to walk to the car?

Answer 1

The ranger should walk 0.577 miles per hour in order to decrease the time required to reach the car.

Explanation:

Suppose the ranger reaches x miles from the end of the road which becomes the horizontal distance and the vertical distance is 1 miles. A right angle triangle can be obtained that shows the ranger walks along the hypothesis.

`\text {Hypothesis}=\sqrt{x^(2)+1}`

The distance left to reach the car is = 5 - x

`\text { Distance }(\mathrm{D})=\text { speed }(\mathrm{s}) * \text { time }(\mathrm{t})`

`\text { time }=\frac{\text {distance}}{\text {speed}}`

To calculate total time taken, then the function becomes

`T(x)=\frac{\sqrt{x^(2)+1}}{2}+((5-x))/(4)`

In order to find the minimized time, differentiate the function T as follows

`(d)/(d x T(x))=(d)/(d x)\left(\left(\frac{\sqrt{x^(2)+1}}{2}\right)+\left(((5-x))/(4)\right)\right)`

`(d)/(d x T(x))=(1)/(2)\left(\frac{2 x}{2 \sqrt{x^(2)+1}}\right)+\left((-1)/(4)\right)`

`(d)/(d x T(x))=\frac{x}{2 \sqrt{x^(2)+1}}-(1)/(4)`

Equate the derivative to zero and obtained

`\frac{x}{2 \sqrt{x^(2)+1}}-(1)/(4)=0`

`\frac{x}{2 \sqrt{x^(2)+1}}=(1)/(4)`

`\frac{x}{\sqrt{x^(2)+1}}=(2)/(4)`

`\frac{x}{\sqrt{x^(2)+1}}=(1)/(2)`

Squaring both sides

`(x^(2))/(x^(2)+1)=(1)/(4)`

`4 x^(2)=x^(2)+1`

`4 x^(2)-x^(2)=1`

`3 x^(2)=1`

`x^(2)=(1)/(3)`

`x=\sqrt{(1)/(3)}`

The ranger should walk 0.577 miles per hour in order to decrease the time required to reach the car.