1. ## Logarithms

The concentration of alcohol $C$% in John's blood when he consumes alcohol could be represented by $C=0.3te^{-0.912t}$ where $t$ is the number of hours after he consumes alcohol.

Find the value of $t$ when the concentration of alcohol in John's blood is a maximum.

Attempt:

$100=0.3te^{-0.912t}$

$\frac{100}{0.3}=te^{-0.912t}$

$\frac{100}{0.3}=lnt+lne^{-0.912t}$

$\frac{100}{0.3}=lnt-0.912t$

stucked

2. Originally Posted by Punch
The concentration of alcohol $C$% in John's blood when he consumes alcohol could be represented by $C=0.3te^{-0.912t}$ where $t$ is the number of hours after he consumes alcohol.

Find the value of $t$ when the concentration of alcohol in John's blood is a maximum.

Attempt:

$100=0.3te^{-0.912t}$

$\frac{100}{0.3}=te^{-0.912t}$

$\frac{100}{0.3}=lnt+lne^{-0.912t}$

$\frac{100}{0.3}=lnt-0.912t$

stucked
Are you familiar with basic optimization problems using differentiation?

You want to find a value of t that satisfies these two equations: $\frac{dC}{dt} = 0$ and $\frac{d^2C}{dC^2} < 0$

Having found that value of t substitute it into your original equation.

3. Originally Posted by Punch
The concentration of alcohol $C$% in John's blood when he consumes alcohol could be represented by $C=0.3te^{-0.912t}$ where $t$ is the number of hours after he consumes alcohol.

Find the value of $t$ when the concentration of alcohol in John's blood is a maximum.

Attempt:

$100=0.3te^{-0.912t}$

$\frac{100}{0.3}=te^{-0.912t}$

$\color{red}{ln\frac{100}{0.3}}$ $=lnt+lne^{-0.912t}$
But I think he would be dead long before 100%.

4. Originally Posted by Anonymous1
But I think he would be dead long before 100%.
Yes that would be correct, but 100 is not the max.

5. Originally Posted by Gusbob
Are you familiar with basic optimization problems using differentiation?

You want to find a value of t that satisfies these two equations: $\frac{dC}{dt} = 0$ and $\frac{d^2C}{dC^2} < 0$

Having found that value of t substitute it into your original equation.
I have tried that but I got stucked.

$
C=0.3te^{-0.912t}
$

$\frac{dC}{dt} = 0.3t(e^{-0.912t})(-0.912)+e^{-0.912t}(0.3)$

$= -0.912te^{-0.912t}+e^{-0.912t}$

then, $0.912te^{-0.912t} = e^{-0.912t}$

but this seems so wrong

6. Originally Posted by Punch
I have tried that but I got stucked.

$
C=0.3te^{-0.912t}
$

$\frac{dC}{dt} = 0.3t(e^{-0.912t})(-0.912)+e^{-0.912t}(0.3)$

$= -0.912te^{-0.912t}+e^{-0.912t}$

then, $0.912te^{-0.912t} = e^{-0.912t}$

but this seems so wrong
Your differentiation is correct. I got

$\frac{dC}{dt} = 0.3 ( e^{-0.921t} - 0.921t e^{-0.921t})$

You've done well so far, you just need to continue. Exponentials should cancel out -> divide both sides by $e^{-0.912t}$. Just because the number isn't pretty doesn't mean it's wrong.

7. I got it!! 0.912t=1

t=1.1 ^_^ thanks