00:01
In this problem, we're given the concentration of a substance, and then we're asked to find its maximum on the interval from 0 to 12.
00:08
And notice that i've rewritten this function c of t using fractions instead of decimals, and i think it's more helpful this way, but we'll see in a minute.
00:17
So what we'll do is take the derivative of c of t, set it equal to zero, solve for t, and then the value that we get for t will plug that back into c of t.
00:27
So first of all, the derivative is we have 8 times and we'll have negative 2 5ths times e to the same power plus 3 5ths times e to the negative 3 5ths times t.
00:50
Okay, and then i'm going to simplify this a little bit, so i'll pull out the 5 and then i'm also going to pull out an e to the negative 3 5th's t.
01:02
So i'll have 3 and then minus 2, e to the t over 5.
01:12
And the reason why i pulled out the exponential is because this is never equal to 0.
01:18
So if we want the derivative equal to 0, then that means that this term right here must be equal to 0.
01:27
So i have e to the t over 5.
01:31
This equals 3 halves.
01:33
And i can take the natural log of both sides to get t over five equals natural log of three halves.
01:46
So t equals five times the natural log of three halves.
01:54
And you might be wondering whether or not this value falls within this interval.
02:02
And it turns out it does.
02:04
One way to show that without using a calculator is to note that 1 is less than 3 halves, which is less than e.
02:20
And then i can take the natural log of all of these numbers because the natural log is an increasing function.
02:34
Okay...