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

Answer:

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

Step-by-step 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

$\text { Distance }(\mathrm{D})=\text { speed }(\mathrm{s}) \times \text { time }(\mathrm{t})$

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

To calculate total time taken, then the function becomes

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

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

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

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

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

Equate the derivative to zero and obtained

$\frac{x}{2 \sqrt{x^{2}+1}}-\frac{1}{4}=0$

$\frac{x}{2 \sqrt{x^{2}+1}}=\frac{1}{4}$

$\frac{x}{\sqrt{x^{2}+1}}=\frac{2}{4}$

$\frac{x}{\sqrt{x^{2}+1}}=\frac{1}{2}$

Squaring both sides

$\frac{x^{2}}{x^{2}+1}=\frac{1}{4}$

$4 x^{2}=x^{2}+1$

$4 x^{2}-x^{2}=1$

$3 x^{2}=1$

$x^{2}=\frac{1}{3}$

$x=\sqrt{\frac{1}{3}}$

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