Question

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, t>=0.

a) when is the concentration greatest?
b)what is the greatest concentration?
c)what is lowest concentration and when does the lowest concentration occur?

Answer 1

i am confused, is it really
\[C=\frac{3t}{27}+t^3\]?

Answer 2

because this function grows without bound, so the bigger \(t\) is the bigger \(C\) is

Answer 3

are you sure this is the right formula?

Answer 4

C= (3t)/(27+t^3) when t>=0

Answer 5

\[C= \frac{ 3t }{ 27+t^3 }\]

Answer 6

To find the min/max of a function, find when the derivative is 0

Answer 7

first derivative test?

Answer 8

what min and max? u mean local min and local max? or absolute min and absolute max?

Answer 9

Absolute. You need to find the absolute extrema.

Answer 10

after i found the absolute extrema, would that be the greatest concentration?

Answer 11

There's gotta be an absolute max and absolute min according to the question.

Answer 12

how can i find the time of the greatest concentration?

Answer 13

Find the first and second derivative. Then find the stationary points.
\[C=\frac{ 3t}{ 27+t^3 }\]
\[C'=\frac{3(27+t^3)-3t^2(3t)}{(27+t^3)^2}\]
\[=\frac{81-6t^3}{(27+t^3)^2}\]
\[C"=\frac{-18t^2(27+t^3)^2-6t^2(27+t^3)(81-6t^3)}{(27+t^3)^4}\]
\[=\frac{-486t^2-18t^5-486t^2+36t^5}{(27+t^3)^3}\]
\[=\frac{-972t^2+18t^5}{(27+t^3)^3}\]
At S.P's(Stationary Points) C'=0
\[81=6t^3\]
\[t^3=\frac{27}{2}\]
\[t=\frac{3}{\sqrt[3]{2}}\]
\[\huge C=\frac{ \frac { 9 }{ \sqrt[3]{2} } }{ 27+\frac{ 27 }{ 2 } } \]
\[C=\frac{2}{9\sqrt[3]{2}}\]

Answer 14

\[C''<0\]
Therefore maximum.

Answer 15

So I did a) and b) for you already.