We are given the distance of the shoreline (15ft) and distance from shore (30ft). The optimal point for the pooch to switch from running to swimming lies at some point x between 0ft and 15ft along the shoreline. We can divide the shoreline into two segments, x and 15 less x.
\[ \begin{align} shore &= 15 = x + (15-x) \\ water &= 30 \\ hypoteneuse &= \sqrt{x^{2}+30^{2}} \end{align} \]
The total time is a composite of swimtime on the hypoteneuse of the triangle with base of distance x and height of distance 30; and runtime on distance of 15 less x.
\[ \begin{align} time &= swimtime + runtime \\ &= \frac{(15-x)ft}{22 ft/sec} + \frac{(\sqrt{x^{2}+30^{2})ft}}{1.5ft/sec} \\ &= \frac{15}{22}-\frac{x}{22}+\frac{\sqrt{x^{2}+30^{2}}}{1.5} \end{align} \]
The endpoints of the interval are unlikely to provide minima, but we check to be sure.
\[ \begin{align} c(0) &= \frac{15}{22}-\frac{0}{22}+\frac{\sqrt{0^{2}+30^{2}}}{1.5} \\ &= \frac{15}{22}+\frac{30}{1.5} \\ &= 20.68182 \\ c(15) &= \frac{15}{22}-\frac{15}{22}+\frac{\sqrt{15^{2}+30^{2}}}{1.5} \\ &= \frac{\sqrt{1125}}{1.5} \\ &= 22.36068 \end{align} \]
The critical values are found by solving for \(c'(0)\).
\[ \begin{align} c'(x)=0-\frac{1}{22}&+\frac{x}{1.5\sqrt{x^{2}+30^{2}}} \\ \frac{1}{22}&=\frac{x}{1.5\sqrt{x^{2}+30^{2}}} \\ \sqrt{x^{2}+30^{2}}&=22x \\ x^{2}+30^{2}&=(22x)^{2} \\ 483x^{2}&=900 \\ x^{2}&=\frac{900}{483} \\ x&=1.365047 \\ \end{align} \]
This critical value designated a minimum on the [0, 15] interval, and we can solve for swimming and running distance and time.
shoredist_swim <- sqrt(900/483)
shoredist_run <- 15-shoredist_swim
waterdist <- 30
hypoteneuse <- sqrt(waterdist^2 + shoredist_swim^2)
runtime <- shoredist_run / 22
swimtime <- hypoteneuse / 1.5
shoredist_run## [1] 13.63495hypoteneuse## [1] 30.03104runtime## [1] 0.6197706swimtime## [1] 20.02069To minimize time-to-stick-retrieval, the dog should run along 13.6ft of shoreline and then swim 33ft through the water. That’s about .62sec of running followed by nearly 20sec of swimming. Good dog!