Difficult Math Problem. Can anyone solve it?

Question

A severely ill patient is given an intramuscular injection of a highly potent antibiotic agent. It is known that the blood concentration, C, measured in micrograms per cc of the agent, increases according to the formula C = 2t/(t^2 + 9), where t is measured in hours after the time of injection. Correct to the nearest tenth of an hour, when will the concentration be maximal?

Answer 1

C' = [(t^2 + 9)(2) - (2t)(2t)]/(t^2 +9)^2
= [2t^2 + 18 - 4t^2]/(t^2+9)^2
= [18 - 2t^2]/(t^2 +9)^2
Set the numerator equal to zero.
18 - 2t^2 = 0
18 = 2t^2
9 = t^2
t = -3 or 3 (discard the negative value because time is not negative).
t = 3

Answer 2

You want to maximize the function
> C(t) = 2t/(t^2 + 9)
This function is continuous and differentiable, so its maximum will be when the derivative C'(t) = 0. The derivative can be found using the chain rule:
> C'(t) = [2(t^2 + 9) - (2t)(2t)] / [(t^2 + 9)^2]
> C'(t) = [18 - 2t^2] / [...] = 0
This fraction is zero whenever the numerator is zero, so
> 18 - 2t^2 = 0
> t^2 = 9
> t = sqrt 9 = +/- 3 (only positive solution makes sense)

So you find 3.0 hours; the concentration at that time is 1/3 mg/cc.

(It is easy to check that this is a maximum, not a minimum.)

Answer 3

A maximum is found by taking the derivative of the function.

Use the quotient rule(check the link below for the math of it)


C'(t) = [(t^2 + 9) * 2 - (2t * 2t) ] / (t^2 + 9)^2
= (2t^2 + 18 - 4t^2) / (t^4 +18t^2 + 81)

Set C'(t) = 0, to find the values of t, which give you a maximum concentration:

C'(t) = 0 = (18 - 2t^2) / (t^4 +18t^2 + 81)
0 * (t^4 +18t^2 + 81) = (18 - 2t^2)
0 = (18 - 2t^2)
2t^2 = 18
t = 3 or -3

Since time cannot be negative, t =3 hours.

Answer 4

answer: t=3

find the value for which the derivative of C is equal to zero.
So, C' = (2(t^2 + 9) - 2t(2t))/(t^2+9)^2 = (-2t^2 + 18)/(t^2+9)^2 = 0.
C' = 0 when t=3. note: t = -3 is not is another solution but should be discarded because of the context of the question.

Answer 5

maximum deals with deriviteve, so c'=(-2t^2+18)/(t^2 + 9)^2
set c'=0 solve for t
-2t^2+18=0, t=+or - 3 since time is always positive
t=3 is the choice to check if it is a max you could take the second derivitive or analyze c'
if c' is less than 3 c' is positive and c is increasing
if c' is greater than 3 c' is negetive and c is decreasing
so at t=3hours c has its maximum
good luck!

Answer 6

I betchya trying out ur homework..jez say so n skip da 1st 2 lines.

Differential Eqn:
Let u=2t
v=t^2 + 9
dC/dt = d(u/v)x
= (v*du/dx - dv/dx*u)/oh god I'm sorry but I think I 4got the formula. Still for all equate ur answer to zero and that willl be the time when the rate of change in concn is 0 and hence the concn has 2 be a max(or min, but that's not the q) Also it's maximum. Not maximal.

Answer 7

Yep the guy is Perelman yet he did not quit his activity to artwork in this subject. He grew to become right into a traveling professor at UC Berkeley a jointly as in the past and greater at the instant he labored in a Russian institute in St. Petersburg. He very at the instant quit his activity as a results of his dissatisfaction with the mathematical community's loss of a solid reaction to what he considers an attempt to downplay his contributions to the evidence via yet another universal mathematician named Yau. As for the Nobel of arithmetic, yep Fields medal is in simple terms that and slightly greater, however the reason of the shortcoming of a Nobel for arithmetic has not something to do along with his spouse cheating Nobel. truthfully Nobel did not have a spouse. maximum historians have faith that Nobel in simple terms did not locate arithmetic clever sufficient to commit an award to.

Answer 8

Right around 3 hours, with a concentration of .333 ug/cc

Answer 9

3.5 hours

To the nearest TENTH............



t3.4 = .33c
t3.5 = .57c
t3.6 = .32c