Find the exact curve where the following curve is the steepest:
y= 50/ (1+6e^-2t) for t> or =to 0

First take the derivative of the Y w.r.t t
so
dy/dt =50(-1) (1 + 6e^(-2t))^(-2) . (6e^(-2t) . (-2)
=600 e^(-2t) / (1 + 6e^(-2t))^2
to find that where it is maximum we take the drivative again
d²y/dt² = 600 e^(-2t) . (-2) . (1 + 6e^(-2t))^(-2) + 600 e^(-2t) . (-2) (1 + 6e^(-2t))^(-3) . (-12 e^(-2t)) = [-1200 e^(-2t) (1 + 6e^(-2t)) + 1200 e^(-2t) . 12e^(-2t)] / (1 + 6e^(-2t))^3 = 1200 e^(-2t) [6e^(-2t) - 1] / (1 + 6e^(-2t))^3 which will be 0 when 6e^(-2t) = 1, so e^(-2t) = 1/6 => -2t = ln (1/6) = -ln 6 => t = (1/2) ln 6 
for value of y we know that
6e^(-2t) = 1, so y = 50 / (1 + 6e^(-2t)) = 50 / (1 + 1) = 25. So the coordinates are ((ln 6) / 2, 25).

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