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enot [183]
3 years ago
9

Brandon is on one side of a river that is 50 m wide and wants to reach a point 300 m downstream on the opposite side as quickly

as possible by swimming diagonally across the river and then running the rest of the way. find the minimum amount of time if brandon can swim at 2 m/s and run at 5 m/s
Mathematics
1 answer:
AlexFokin [52]3 years ago
8 0
Let P be Brandon's starting point and Q be the point directly across the river from P. 
<span>Now let R be the point where Brandon swims to on the opposite shore, and let </span>
<span>QR = x. Then he will swim a distance of sqrt(50^2 + x^2) meters and then run </span>
<span>a distance of (300 - x) meters. Since time = distance/speed, the time of travel T is </span>

<span>T = (1/2)*sqrt(2500 + x^2) + (1/6)*(300 - x). Now differentiate with respect to x: </span>

<span>dT/dx = (1/4)*(2500 + x^2)^(-1/2) *(2x) - (1/6). Now to find the critical points set </span>
<span>dT/dx = 0, which will be the case when </span>

<span>(x/2) / sqrt(2500 + x^2) = 1/6 ----> </span>

<span>3x = sqrt(2500 + x^2) ----> </span>

<span>9x^2 = 2500 + x^2 ----> 8x^2 = 2500 ---> x^2 = 625/2 ---> x = (25/2)*sqrt(2) m, </span>

<span>which is about 17.7 m downstream from Q. </span>

<span>Now d/dx(dT/dx) = 1250/(2500 + x^2) > 0 for x = 17.7, so by the second derivative </span>
<span>test the time of travel, T, is minimized at x = (25/2)*sqrt(2) m. So to find the </span>
<span>minimum travel time just plug this value of x into to equation for T: </span>

<span>T(x) = (1/2)*sqrt(2500 + x^2) + (1/6)*(300 - x) ----> </span>

<span>T((25/2)*sqrt(2)) = (1/2)*(sqrt(2500 + (625/2)) + (1/6)*(300 - (25/2)*sqrt(2)) = 73.57 s.</span><span>
</span><span>
</span><span>
</span><span>
</span><span>mind blown</span>
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