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>
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</span><span>mind blown</span>
Use
bodmas
multiply 6 and 3= 18
then add the answer with 23
18+23=455
the answer you get, subtract it from 7
455-7=448
the answer is 448
