Technicians working for the Ministry of Natural Resources have found that the amount of a pollutant in a certain river can be represented by , , where is the time (in years) since a clean-up campaign started. At what time was the pollutant at its lowest level?
Therefore, the pollution in 0.000240991 year was at its lowest level of 0.963
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Textbook answer: 0.049 years
Did I take the first derivative wrong? I checked it and I don't see anything wrong with it. Please show me my mistake and fix it for me?
P(t) = 2t + 1/(162t + 1)
*use quotient rule on second term*
P'(t) = 2 - 162/(162t + 1)^2
*minimum occurs when P'(t) = 0*
2 - 162/(162t + 1)^2 = 0
2 = 162/(162t + 1)^2
2(162t + 1)^2 = 162
2(26244t^2 + 324t + 1) = 162
52488t^2 + 648t -160 = 0
t = 0.04938 or t = -0.06173
Cannot be negative, therefore t = 0.04938