College Algebra, Chapter 1, 1.5, Section 1.5, Problem 84

Suppose that a boardwalk is parallel to and $210 ft$ in land from a straight shoreline where beach lies between the boardwalk and the shoreline. If a man is standing on the boardwalk, exactly $750ft$ across the sand from his beach umbrella, which is right at the shoreline, and he walks $4 ft/s$ on the boardwalk and $2ft$ on the sand. How far should he walk on the boardwalk before veering off onto the sand if he wishes to reach his umbrella in exactly $4 min 45 s$?

If we let $b$ be the distance up to the point directly below the umbrella, then by Pythagorean Theorem,


$\begin{equation} \begin{aligned} 750^2 =& b^2 + 210^2 \\ \\ b^2 =& 750^2 - 210^2 \\ \\ b =& 720 ft \end{aligned} \end{equation}$


If we let $720 - x$ be the distance that the man will take on boardwalk, then by Pythagorean Theorem on the right side,








$\begin{equation} \begin{aligned} z^2 =& 210^2 + x^2 \\ \\ z =& \sqrt{x^2 + 210^2} ft \end{aligned} \end{equation}$


Now, the total of travel is the sum of the time it takes a man to travel in boardwalk and the time it takes a man to walk in sand. Thus,


$\begin{equation} \begin{aligned} t_T =& t_1 + t_2 && \text{Model, recall that } t = \frac{d}{v} \\ \\ 4 min \left( \frac{60 s}{1 min} \right) + 45 s =& \frac{720 - x}{4} + \frac{\sqrt{x^2 + 210^2}}{2} && \\ \\ 285 =& \frac{720 - x}{4} + \frac{\sqrt{x^2 + 210^2}}{2} && \text{Multiply both sides by } 4 \\ \\ 1140 =& 720 - x + 2 \sqrt{x^2 + 210^2} && \text{Add $x$ and subtract $720m$ both sides of the equation} \\ \\ x - 420 =& 2 \sqrt{x^2 + 210^2} && \text{Square both sides} \\ \\ (x - 420)^2 =& 4(x^2 + 210^2) && \text{Expand} \\ \\ x^2 - 840x + 420^2 =& 4x^2 + 4(210^2) && \text{Combine like terms} \\ \\ 3x^2 + 840x =& 0 && \text{Factor out } 3x \\ \\ 3x(x + 280) =& 0 && \text{Zero Product Property} \\ \\ x =& 0 \text{ and } x + 280 = 0 && \text{Solve for } x \\ \\ x =& 0 \text{ and } x = -280 && \end{aligned} \end{equation}$


Thus, if we substitute $x$ with the distance required, then


$\begin{equation} \begin{aligned} d_1 =& 720 - x = 720 - 0 = 720ft \\ \\ d_2 =& 720 - (-280) = 1000 ft \text{ (absurd)} \end{aligned} \end{equation}$


It shows that in order for the man to reach his umbrella by $4 min$ and $45 sec$, he must take the boardwalk distance of $720ft$ then take a $210 ft$ length by sand.