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.