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) When is the concentration greatest?

Answer 1

Answer: 2.381 hours

Answer 2

@yun2thejae

Answer 3

Oh I guess it is asking you to find the absolute maximum of that graph. give me second. I'm a tad rusty .

Answer 4

Here are the steps:
1. Take the derivative of that equation.
2. Set the derivative equal to zero to find critical numbers.
3. Test the critical numbers to find local maximum points.
4. Plug those maximum points back into the original equation.
5. The largest number is your answer. in this case, 2.381.

Answer 5

I'm going to try to work this out,. but you try it also in case I do it too slow.

Answer 6

u did it already or was that a guess

Answer 7

? you gave me the answer up there. I have not finished it yet, I'm haing a tad of trouble finding the critical numbers to test

Answer 8

Oh wait, I see why I am having problems. give me a second.

Answer 9

k

Answer 10

Take the derivative, you get (81+3t^3-9t^3)/(27+t^3)^2. set that equal to zero. you have to solve for 81+3t^3-9t^3=0

Answer 11

81=6t^3
13.5=t^3
Cube root of 13.5 = t.

Answer 12

if you test that point, it should be the absolute maximum of the equation. cube root of 13.5 equals the answer you gave me up there.

Answer 13

sorry it took so long. I have the lights turned off while trying to hand write math LOL. I kept messing up simple algebra.

Answer 14

i got \[\frac{ 81-6t^3 }{ (27+t^3)^2 }\] for my derivative

Answer 15

is that 81-6t^3? I can't see very well

Answer 16

yeah

Answer 17

yeah that is correct. now set it equal to zero. When you do that, the bottom goes away because you cross multiplied it with the zero. so you end up with 81-6t^3=0

Answer 18

oh okay ty o.O

Answer 19

solve for t and that is your answer. Ideally, you're expected to check to see if it actually a maximum, but in this case there is only one number so..you don't have to. Although, I would on the Ap exam though.

Answer 20

thx :P

Answer 21

You're welcome. Anything else?

Answer 22

yeah

Answer 23

Post it up and i'll see if I can help.