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A park ranger needs to get to a point 12 kilometers downstream on the opposite side of a river that is 2 kilometers wide. If the ranger can row at 5 kilometers per hour and run at 7 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 12 kilometers downstream on the opposite side of a river that is 2 kilometers wide. If the ranger can row at 5 kilometers per hour and run at 7 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]{$2$};
\draw[latex-latex,thick] (0,-.75) -- (10,-.75) node[midway,below]{$12$};
\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]{$12 - x$};
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\end{center}Using the Pythagorean theorem to solve for the distance traveled on the river gives $x^{2} + 4=c^2$. This gives $d_1=\sqrt{x^{2} + 4}$ and the distance traveled on land is $d_2=12 - x$. 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 $t_1=\frac{\sqrt{x^{2} + 4}}{5}$ and the time on land is $t_2=\frac{12 - x}{7}$. The time function is $T(x)=- \frac{x}{7} + \frac{\sqrt{x^{2} + 4}}{5} + \frac{12}{7}$ and it has domain $[0,12]$. Setting $T'(x)=0$ gives the equation $\frac{x}{5 \sqrt{x^{2} + 4}} - \frac{1}{7}=0$. Solving gives the critical number $\frac{5 \sqrt{6}}{6}\approx 2.04$. 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: $T\left(0\right)=\frac{74}{35}\approx 2.11$. $T\left(12\right)=\frac{2 \sqrt{37}}{5}\approx 2.43$. $T\left(\frac{5 \sqrt{6}}{6}\right)=\frac{4 \sqrt{6}}{35} + \frac{12}{7}\approx 1.99$. The minimum occurs at $\left( \frac{5 \sqrt{6}}{6}, \ \frac{4 \sqrt{6}}{35} + \frac{12}{7}\right)$. }
\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 12 kilometers downstream on the opposite side of a river that is 2 kilometers wide. If the ranger can row at 5 kilometers per hour and run at 7 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} + 4=c^2 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%204%3Dc%5E2%20" alt="LaTeX: \displaystyle x^{2} + 4=c^2 " data-equation-content=" \displaystyle x^{2} + 4=c^2 " /> . This gives <img class="equation_image" title=" \displaystyle d_1=\sqrt{x^{2} + 4} " src="/equation_images/%20%5Cdisplaystyle%20d_1%3D%5Csqrt%7Bx%5E%7B2%7D%20%2B%204%7D%20" alt="LaTeX: \displaystyle d_1=\sqrt{x^{2} + 4} " data-equation-content=" \displaystyle d_1=\sqrt{x^{2} + 4} " /> and the distance traveled on land is <img class="equation_image" title=" \displaystyle d_2=12 - x " src="/equation_images/%20%5Cdisplaystyle%20d_2%3D12%20-%20x%20" alt="LaTeX: \displaystyle d_2=12 - x " data-equation-content=" \displaystyle d_2=12 - 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} + 4}}{5} " src="/equation_images/%20%5Cdisplaystyle%20t_1%3D%5Cfrac%7B%5Csqrt%7Bx%5E%7B2%7D%20%2B%204%7D%7D%7B5%7D%20" alt="LaTeX: \displaystyle t_1=\frac{\sqrt{x^{2} + 4}}{5} " data-equation-content=" \displaystyle t_1=\frac{\sqrt{x^{2} + 4}}{5} " /> and the time on land is <img class="equation_image" title=" \displaystyle t_2=\frac{12 - x}{7} " src="/equation_images/%20%5Cdisplaystyle%20t_2%3D%5Cfrac%7B12%20-%20x%7D%7B7%7D%20" alt="LaTeX: \displaystyle t_2=\frac{12 - x}{7} " data-equation-content=" \displaystyle t_2=\frac{12 - x}{7} " /> . The time function is <img class="equation_image" title=" \displaystyle T(x)=- \frac{x}{7} + \frac{\sqrt{x^{2} + 4}}{5} + \frac{12}{7} " src="/equation_images/%20%5Cdisplaystyle%20T%28x%29%3D-%20%5Cfrac%7Bx%7D%7B7%7D%20%2B%20%5Cfrac%7B%5Csqrt%7Bx%5E%7B2%7D%20%2B%204%7D%7D%7B5%7D%20%2B%20%5Cfrac%7B12%7D%7B7%7D%20" alt="LaTeX: \displaystyle T(x)=- \frac{x}{7} + \frac{\sqrt{x^{2} + 4}}{5} + \frac{12}{7} " data-equation-content=" \displaystyle T(x)=- \frac{x}{7} + \frac{\sqrt{x^{2} + 4}}{5} + \frac{12}{7} " /> and it has domain <img class="equation_image" title=" \displaystyle [0,12] " src="/equation_images/%20%5Cdisplaystyle%20%5B0%2C12%5D%20" alt="LaTeX: \displaystyle [0,12] " data-equation-content=" \displaystyle [0,12] " /> . 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}{5 \sqrt{x^{2} + 4}} - \frac{1}{7}=0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bx%7D%7B5%20%5Csqrt%7Bx%5E%7B2%7D%20%2B%204%7D%7D%20-%20%5Cfrac%7B1%7D%7B7%7D%3D0%20" alt="LaTeX: \displaystyle \frac{x}{5 \sqrt{x^{2} + 4}} - \frac{1}{7}=0 " data-equation-content=" \displaystyle \frac{x}{5 \sqrt{x^{2} + 4}} - \frac{1}{7}=0 " /> . Solving gives the critical number <img class="equation_image" title=" \displaystyle \frac{5 \sqrt{6}}{6}\approx 2.04 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B5%20%5Csqrt%7B6%7D%7D%7B6%7D%5Capprox%202.04%20" alt="LaTeX: \displaystyle \frac{5 \sqrt{6}}{6}\approx 2.04 " data-equation-content=" \displaystyle \frac{5 \sqrt{6}}{6}\approx 2.04 " /> . 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{74}{35}\approx 2.11 " src="/equation_images/%20%5Cdisplaystyle%20T%5Cleft%280%5Cright%29%3D%5Cfrac%7B74%7D%7B35%7D%5Capprox%202.11%20" alt="LaTeX: \displaystyle T\left(0\right)=\frac{74}{35}\approx 2.11 " data-equation-content=" \displaystyle T\left(0\right)=\frac{74}{35}\approx 2.11 " /> . <img class="equation_image" title=" \displaystyle T\left(12\right)=\frac{2 \sqrt{37}}{5}\approx 2.43 " src="/equation_images/%20%5Cdisplaystyle%20T%5Cleft%2812%5Cright%29%3D%5Cfrac%7B2%20%5Csqrt%7B37%7D%7D%7B5%7D%5Capprox%202.43%20" alt="LaTeX: \displaystyle T\left(12\right)=\frac{2 \sqrt{37}}{5}\approx 2.43 " data-equation-content=" \displaystyle T\left(12\right)=\frac{2 \sqrt{37}}{5}\approx 2.43 " /> . <img class="equation_image" title=" \displaystyle T\left(\frac{5 \sqrt{6}}{6}\right)=\frac{4 \sqrt{6}}{35} + \frac{12}{7}\approx 1.99 " src="/equation_images/%20%5Cdisplaystyle%20T%5Cleft%28%5Cfrac%7B5%20%5Csqrt%7B6%7D%7D%7B6%7D%5Cright%29%3D%5Cfrac%7B4%20%5Csqrt%7B6%7D%7D%7B35%7D%20%2B%20%5Cfrac%7B12%7D%7B7%7D%5Capprox%201.99%20" alt="LaTeX: \displaystyle T\left(\frac{5 \sqrt{6}}{6}\right)=\frac{4 \sqrt{6}}{35} + \frac{12}{7}\approx 1.99 " data-equation-content=" \displaystyle T\left(\frac{5 \sqrt{6}}{6}\right)=\frac{4 \sqrt{6}}{35} + \frac{12}{7}\approx 1.99 " /> . The minimum occurs at <img class="equation_image" title=" \displaystyle \left( \frac{5 \sqrt{6}}{6}, \ \frac{4 \sqrt{6}}{35} + \frac{12}{7}\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%20%5Cfrac%7B5%20%5Csqrt%7B6%7D%7D%7B6%7D%2C%20%5C%20%20%5Cfrac%7B4%20%5Csqrt%7B6%7D%7D%7B35%7D%20%2B%20%5Cfrac%7B12%7D%7B7%7D%5Cright%29%20" alt="LaTeX: \displaystyle \left( \frac{5 \sqrt{6}}{6}, \ \frac{4 \sqrt{6}}{35} + \frac{12}{7}\right) " data-equation-content=" \displaystyle \left( \frac{5 \sqrt{6}}{6}, \ \frac{4 \sqrt{6}}{35} + \frac{12}{7}\right) " /> . </p> </p>