Question

A boardwalk is parallel to and 210 ft inland from a straight shoreline. A sandy beach lies between the boardwalk and the shoreline. A man is standing on the boardwalk, exactly 750 ft across the sand from his beach umbrella, which is right at the shoreline. The man walks 4 ft/s on the boardwalk and 2 ft/s 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?

Short answer

The man could walk either 440 ft or 720 ft on boardwalk.

Step-by-step solution

Step-1. Determine the variables.

Consider the figure shown below:

Let x be the distance of boardwalk the man doesn’t walk.

Now apply the Pythagoras theorem and find the value of b.

$\begin{array}{l}{750}^2=b^2+{210}^2\\ b=\sqrt{518400}\\ b=720\end{array}$

Again, solve for h.

$\begin{array}{l}{h}^2={210}^2+x^2\\ h=\sqrt{44100+x^2}\end{array}$

Step-2. Setup model.

We know that $\mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}}$.

The given time is 4 min 45 sec or 285 seconds.

Since, the maximum time is 285 seconds, therefore,

$\begin{array}{l}\frac{720-x}{4}+\frac{\sqrt{44100+x^2}}{2}=285\\ 720-x+2\sqrt{44100+x^2}=4\left(285\right)\\ 2\sqrt{44100+x^2}=420+x\\ 4(44100+x^2)=x^2+840x+176400\\ 3x^2-840x=0\end{array}$

Step-3. Solve for x.

$\begin{array}{l}3x^2-840x=0\\ x(3-840x)=0\\ x=0\mathrm{or}x=280\end{array}$

We know that x is the distance of the boardwalk the man doesn’t walk.

It means$x=0$represents man doesn’t walk on the sand he walks all 720 ft on the boardwalk.

Also,$x=280$represents man doesn’t walk 280 ft on the boardwalk it means he must walk $720-280=440$ft on the boardwalk.

Hence, the man could walk either 440 ft or 720 ft on the boardwalk.