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AP Calculus: The concentration C of a certain chemical in the bloodstream t hours after injection into muscle tissue is given by C=(3t)/(27+t^3) When is the concentration greatest?
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take the derivate and set it equal to 0
k, i got \[\frac{ 81-6t }{ (27+t^3)^2 }\]
@TweT226
so set that = 0 and solve for t
how
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but it should be 81 - 6t^3
$$C(t)=\frac{3t}{27+t^3}\\C'(t)=\frac{3(27+t^3)-3t(3t^2)}{(27+t^3)^2}=\frac{81+3t^3-9t^3}{(27+t^3)^2}=\frac{81-6t^3}{(27+t^3)^2}\\C'(t_*)=0\\\frac{81-6t_*^3}{(27+t_*^3)^2}=0\\81-6t_*^3=0\\81=6t_*^3\\\frac{27}2=t_*^3\\t_*=\frac3{\sqrt[3]{2}}$$
ooh ok ty ^^
can u help me with another pls? :)
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