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A park ranger needs to get to a point 13 kilometers downstream on the opposite side of a river that is 3 kilometers wide. If the ranger can row at 3 kilometers per hour and run at 11 kilometers hour, how far down stream should the ranger land their boat on the opposite side of the river to minimize their travel time?
Drawing a diagram gives
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\begin{question}A park ranger needs to get to a point 13 kilometers downstream on the opposite side of a river that is 3 kilometers wide. If the ranger can row at 3 kilometers per hour and run at 11 kilometers hour, how far down stream should the ranger land their boat on the opposite side of the river to minimize their travel time?
\soln{9cm}{Drawing a diagram gives\newline\begin{center}
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\draw[color=blue,thick] (-.5,2)--(10.5,2);
\draw[color=blue,thick] (-.5,0)--(10.5,0);
\draw[fill=black] (0,2.25) circle (2pt) node[above]{$A$};
\draw[fill=black] (10,-.25) circle (2pt) node[right]{$B$};
\draw[latex-latex, thick] (0,2)--(0,0) node[midway,left]{$3$};
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\filldraw[thick,color=red] (0,2) -- (6,0) circle (2pt);
\draw[latex-latex] (0,-.5) -- (6,-.5) node[above, midway]{$x$};
\draw[latex-latex] (6,-.5) -- (10,-.5) node[above, midway]{$13 - x$};
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\end{center}Using the Pythagorean theorem to solve for the distance traveled on the river gives $x^{2} + 9=c^2$. This gives $d_1=\sqrt{x^{2} + 9}$ and the distance traveled on land is $d_2=13 - x$. Using $d=rt$ and solving for time gives $t=\frac{d}{r}$. That is, the time is equal to distance divided by speed. The time on the river is $t_1=\frac{\sqrt{x^{2} + 9}}{3}$ and the time on land is $t_2=\frac{13 - x}{11}$. The time function is $T(x)=- \frac{x}{11} + \frac{\sqrt{x^{2} + 9}}{3} + \frac{13}{11}$ and it has domain $[0,13]$. Setting $T'(x)=0$ gives the equation $\frac{x}{3 \sqrt{x^{2} + 9}} - \frac{1}{11}=0$. Solving gives the critical number $\frac{9 \sqrt{7}}{28}\approx 0.850$. By the extreme value theorem the absolute minimumum must be at the endpoints of the domain or the critical numbers. Testing each of the values gives: $T\left(0\right)=\frac{24}{11}\approx 2.18$. $T\left(13\right)=\frac{\sqrt{178}}{3}\approx 4.45$. $T\left(\frac{9 \sqrt{7}}{28}\right)=\frac{4 \sqrt{7}}{11} + \frac{13}{11}\approx 2.14$. The minimum occurs at $\left( \frac{9 \sqrt{7}}{28}, \ \frac{4 \sqrt{7}}{11} + \frac{13}{11}\right)$. }
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>A park ranger needs to get to a point 13 kilometers downstream on the opposite side of a river that is 3 kilometers wide. If the ranger can row at 3 kilometers per hour and run at 11 kilometers hour, how far down stream should the ranger land their boat on the opposite side of the river to minimize their travel time? </p> </p>
<p> <p>Drawing a diagram gives<br><center>
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</center>Using the Pythagorean theorem to solve for the distance traveled on the river gives <img class="equation_image" title=" \displaystyle x^{2} + 9=c^2 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%209%3Dc%5E2%20" alt="LaTeX: \displaystyle x^{2} + 9=c^2 " data-equation-content=" \displaystyle x^{2} + 9=c^2 " /> . This gives <img class="equation_image" title=" \displaystyle d_1=\sqrt{x^{2} + 9} " src="/equation_images/%20%5Cdisplaystyle%20d_1%3D%5Csqrt%7Bx%5E%7B2%7D%20%2B%209%7D%20" alt="LaTeX: \displaystyle d_1=\sqrt{x^{2} + 9} " data-equation-content=" \displaystyle d_1=\sqrt{x^{2} + 9} " /> and the distance traveled on land is <img class="equation_image" title=" \displaystyle d_2=13 - x " src="/equation_images/%20%5Cdisplaystyle%20d_2%3D13%20-%20x%20" alt="LaTeX: \displaystyle d_2=13 - x " data-equation-content=" \displaystyle d_2=13 - x " /> . Using <img class="equation_image" title=" \displaystyle d=rt " src="/equation_images/%20%5Cdisplaystyle%20d%3Drt%20" alt="LaTeX: \displaystyle d=rt " data-equation-content=" \displaystyle d=rt " /> and solving for time gives <img class="equation_image" title=" \displaystyle t=\frac{d}{r} " src="/equation_images/%20%5Cdisplaystyle%20t%3D%5Cfrac%7Bd%7D%7Br%7D%20" alt="LaTeX: \displaystyle t=\frac{d}{r} " data-equation-content=" \displaystyle t=\frac{d}{r} " /> . That is, the time is equal to distance divided by speed. The time on the river is <img class="equation_image" title=" \displaystyle t_1=\frac{\sqrt{x^{2} + 9}}{3} " src="/equation_images/%20%5Cdisplaystyle%20t_1%3D%5Cfrac%7B%5Csqrt%7Bx%5E%7B2%7D%20%2B%209%7D%7D%7B3%7D%20" alt="LaTeX: \displaystyle t_1=\frac{\sqrt{x^{2} + 9}}{3} " data-equation-content=" \displaystyle t_1=\frac{\sqrt{x^{2} + 9}}{3} " /> and the time on land is <img class="equation_image" title=" \displaystyle t_2=\frac{13 - x}{11} " src="/equation_images/%20%5Cdisplaystyle%20t_2%3D%5Cfrac%7B13%20-%20x%7D%7B11%7D%20" alt="LaTeX: \displaystyle t_2=\frac{13 - x}{11} " data-equation-content=" \displaystyle t_2=\frac{13 - x}{11} " /> . The time function is <img class="equation_image" title=" \displaystyle T(x)=- \frac{x}{11} + \frac{\sqrt{x^{2} + 9}}{3} + \frac{13}{11} " src="/equation_images/%20%5Cdisplaystyle%20T%28x%29%3D-%20%5Cfrac%7Bx%7D%7B11%7D%20%2B%20%5Cfrac%7B%5Csqrt%7Bx%5E%7B2%7D%20%2B%209%7D%7D%7B3%7D%20%2B%20%5Cfrac%7B13%7D%7B11%7D%20" alt="LaTeX: \displaystyle T(x)=- \frac{x}{11} + \frac{\sqrt{x^{2} + 9}}{3} + \frac{13}{11} " data-equation-content=" \displaystyle T(x)=- \frac{x}{11} + \frac{\sqrt{x^{2} + 9}}{3} + \frac{13}{11} " /> and it has domain <img class="equation_image" title=" \displaystyle [0,13] " src="/equation_images/%20%5Cdisplaystyle%20%5B0%2C13%5D%20" alt="LaTeX: \displaystyle [0,13] " data-equation-content=" \displaystyle [0,13] " /> . Setting <img class="equation_image" title=" \displaystyle T'(x)=0 " src="/equation_images/%20%5Cdisplaystyle%20T%27%28x%29%3D0%20" alt="LaTeX: \displaystyle T'(x)=0 " data-equation-content=" \displaystyle T'(x)=0 " /> gives the equation <img class="equation_image" title=" \displaystyle \frac{x}{3 \sqrt{x^{2} + 9}} - \frac{1}{11}=0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bx%7D%7B3%20%5Csqrt%7Bx%5E%7B2%7D%20%2B%209%7D%7D%20-%20%5Cfrac%7B1%7D%7B11%7D%3D0%20" alt="LaTeX: \displaystyle \frac{x}{3 \sqrt{x^{2} + 9}} - \frac{1}{11}=0 " data-equation-content=" \displaystyle \frac{x}{3 \sqrt{x^{2} + 9}} - \frac{1}{11}=0 " /> . Solving gives the critical number <img class="equation_image" title=" \displaystyle \frac{9 \sqrt{7}}{28}\approx 0.850 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B9%20%5Csqrt%7B7%7D%7D%7B28%7D%5Capprox%200.850%20" alt="LaTeX: \displaystyle \frac{9 \sqrt{7}}{28}\approx 0.850 " data-equation-content=" \displaystyle \frac{9 \sqrt{7}}{28}\approx 0.850 " /> . By the extreme value theorem the absolute minimumum must be at the endpoints of the domain or the critical numbers. Testing each of the values gives: <img class="equation_image" title=" \displaystyle T\left(0\right)=\frac{24}{11}\approx 2.18 " src="/equation_images/%20%5Cdisplaystyle%20T%5Cleft%280%5Cright%29%3D%5Cfrac%7B24%7D%7B11%7D%5Capprox%202.18%20" alt="LaTeX: \displaystyle T\left(0\right)=\frac{24}{11}\approx 2.18 " data-equation-content=" \displaystyle T\left(0\right)=\frac{24}{11}\approx 2.18 " /> . <img class="equation_image" title=" \displaystyle T\left(13\right)=\frac{\sqrt{178}}{3}\approx 4.45 " src="/equation_images/%20%5Cdisplaystyle%20T%5Cleft%2813%5Cright%29%3D%5Cfrac%7B%5Csqrt%7B178%7D%7D%7B3%7D%5Capprox%204.45%20" alt="LaTeX: \displaystyle T\left(13\right)=\frac{\sqrt{178}}{3}\approx 4.45 " data-equation-content=" \displaystyle T\left(13\right)=\frac{\sqrt{178}}{3}\approx 4.45 " /> . <img class="equation_image" title=" \displaystyle T\left(\frac{9 \sqrt{7}}{28}\right)=\frac{4 \sqrt{7}}{11} + \frac{13}{11}\approx 2.14 " src="/equation_images/%20%5Cdisplaystyle%20T%5Cleft%28%5Cfrac%7B9%20%5Csqrt%7B7%7D%7D%7B28%7D%5Cright%29%3D%5Cfrac%7B4%20%5Csqrt%7B7%7D%7D%7B11%7D%20%2B%20%5Cfrac%7B13%7D%7B11%7D%5Capprox%202.14%20" alt="LaTeX: \displaystyle T\left(\frac{9 \sqrt{7}}{28}\right)=\frac{4 \sqrt{7}}{11} + \frac{13}{11}\approx 2.14 " data-equation-content=" \displaystyle T\left(\frac{9 \sqrt{7}}{28}\right)=\frac{4 \sqrt{7}}{11} + \frac{13}{11}\approx 2.14 " /> . The minimum occurs at <img class="equation_image" title=" \displaystyle \left( \frac{9 \sqrt{7}}{28}, \ \frac{4 \sqrt{7}}{11} + \frac{13}{11}\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%20%5Cfrac%7B9%20%5Csqrt%7B7%7D%7D%7B28%7D%2C%20%5C%20%20%5Cfrac%7B4%20%5Csqrt%7B7%7D%7D%7B11%7D%20%2B%20%5Cfrac%7B13%7D%7B11%7D%5Cright%29%20" alt="LaTeX: \displaystyle \left( \frac{9 \sqrt{7}}{28}, \ \frac{4 \sqrt{7}}{11} + \frac{13}{11}\right) " data-equation-content=" \displaystyle \left( \frac{9 \sqrt{7}}{28}, \ \frac{4 \sqrt{7}}{11} + \frac{13}{11}\right) " /> . </p> </p>