# Exponential growth problem (at least I think it is)

## Homework Statement

The measure of effectiveness of a sleep deprived student studying for a test on a scale of 0 to 1 is given by the formula M(t) = ln(t^2) / 3t, where 't' is the time in hours that a student spends studying, 0 < t ≤ 4.

Determine the maximum measure of effectiveness of studying and the time at which it is reached.

## The Attempt at a Solution

In all honest I've got no idea what going on here, this was just an easter-egg question our teacher handed out at the end of class for a few extra marks. We haven't covered this in class yet, and I'm trying to absorb as much information as possible from my utterly useless textbook, but it's not going very well. It's all just a jumble of balderdash to me at the moment! lol!

If anyone could help me figure this out I'd really appreciate it! I gotta keep my average at 91% of else my university acceptance gets withdrawn! every mark helps!

Thanks again everyone!

The idea is to maximise M with respect to t.

To do this, you can either plot a graph of M against t, or use differentiation.

Is the derivative

M'(t) = 6 - 3ln(t^2) / 9t^2

?

Is the derivative

M'(t) = 6 - 3ln(t^2) / 9t^2

?

No. Use the quotient rule.

I did, and that's what I got. Did I miss something somewhere?

No, I'm wrong. Sorry. I thought your answer was $$6 - \frac {3ln(t^{2})}{9t^{2}}$$.

diazona
Homework Helper
All you missed were some parentheses:
$$M'(t) = (6 - 3\ln(t^2))\ /\ (9t^2)$$
I checked with Mathematica to make sure it's right.

So now that I've got my trusty derivative. Where do I go from here?

Set it equal to zero and solve for t. You''ll get two answers, but one isn't in your domain.

Mark44
Mentor
From a graph of this function, I see a maximum somewhere between 2.6 and 2.75. BTW, this is not exponential growth, nor even logarithmic growth. I don't think there's a name for this kind of function.

Well I ended up getting t = 2.72, and subbing that into the original equation I get M(t) = 0.245.

So hopefully its correct, and I'll post back and let ya'll know how it went! Thanks again everyone for the help!

Well I ended up getting t = 2.72. So hopefully its correct, and I'll post back and let ya'll know how it went! Thanks again everyone for the help!

You don't have to round. The answer is e.

$$\frac {6 - 3ln(t^{2})}{9t^{2}} = 0$$

$$6- 3ln(t^{2}) = 0$$

$$ln(t^{2}) = 2$$

$$t^{2} = e^{2}$$

$$t = e$$ (since -e is not between 0 and 4)

Last edited:
You can use the second derivative test to check that e is indeed a maximum. Find the second derivative of M(t) and evaluate it at e. If it's less than zero, then e is a maximum.

Last edited: