Answer to Question #133560 in Algebra for Jordyn Stierman

Question

A hiker walks up and down a hill. The hill has a cross section that can be modeled by y = -4/3|x - 600| + 800 where x and y are measured in feet and 0 ≤ x ≤ 1200. How far does the hiker walk?

Answer 1

Answer:

The hiker walked a total distance of $2,000 \space meters$

Solutions

The cross section of the hill is given as

$y = -\dfrac {4}{3} \big|x - 600 \big| + 800$ where $y$ is the vertical distance and $x$ is the horizontal displacement, and $0 \leq x \leq 120.$The given equation for the cross section is a modulus function. Let us assume that $x \space and \space y$ are measured in $meters (m).$

When $x = 0 \space m$ , $y = -\dfrac {4}{3} \big|0-600 \big| + 800$

$= -\dfrac {4}{3} \big|-600 \big| + 800$

$= -\dfrac {4}{3}(600) + 800$

$= -800 + 800$

$= 0 m$

Thus, when $x = 0 m$ the hiker was at the base of the hill.

Also, When $x = 120 m$ ,

$y = -\dfrac {4}{3} \big| 120 - 800 \big| + 800$

$= -\dfrac {4}{3}(600) + 800$

$= -800 + 800$

$= 0 m$

Thus, when $x = 120 m$ the hiker had descended from the hill top back to the base.

The maximum value of $y$ occurs when $\big| x - 600 \big| = 0$ , that is when $x = 600$

$\therefore \space max(y) = 800 m$

Therefore, the hiker ascended the hill with a horizontal displacement of 600m and a vertical displacement of 800m, and descended the hill with the same values.

Distance travelled

= distance up the hill + distance down the hill.

$= 2 × \sqrt {x^2 + y^2}$

$= 2× \sqrt {(600m)^2 + (800m)^2}$

$= 2 × \sqrt {360,000m^2 + 640,000m^2}$

$= 2 × \sqrt {1,000,000m^2}$

$= 2 × 1,000m$

$= 2,000 \space meters$

$\therefore$ the hiker waked a distance of 1,000m up the slope and a distance of 1,000m down the slope giving a total distance walked of 2,000 meters.