I have attempted this optimization (through differentiation) question from the exercises I've received through a self-improvement course. I got stuck while differentiating the equation which is supposed to be the exponential growth model for bacteria growth, so I'm unable to solve for the answer, I am hoping someone can help to guide me through this question as I was and still am not adept at maths.
Here is the question:
The bacteria population in a certain colony is given by
$f(t) = 1000t^2e^{-t}$
where $t$ is time in minutes and $t \ge 0$
Find the time at which the population reaches a maximum.
I've immediately attempted the product rule to try and differentiate, but as I said I got stuck, and I can't solve for $t$ without differentiating (I can equate $\frac{dy}{dx}$ to 0 and then find the maximum). I'm also not sure where the information '$t$ is time in minutes' and $t \ge 0$ comes into play when solving.
Please let me know of the steps needed to complete the question and where I've gone wrong, I would love to learn, thanks!
I have attached two screenshots of the full problem, and my attempt at another screenshot.
Best Answer
I can't really follow your handwriting or solution. It says how to solve the problem in the question with a derivative test.
$$ f(t) = 1000t^{2} e^{-t} \tag{1} $$
the product rule is given as
$$ (g \cdot h)^{'} = g^{'} \cdot h+ g \cdot h^{'} \tag{2}$$
the constant doesn't matter here, it simply scales so instead look at
$$ f_{1}(t) = g \cdot h = t^{2} e^{-t} \tag{3}$$
then we have
$$ g(t) = t^{2} , h(t) = e^{-t} \implies g^{'}(t) = 2 t , h^{'}(t) = -e^{-t} \tag{4}$$
which gives us
$$ f_{1}^{'}(t) = 2 t \cdot e^{-t} - t^{2}e^{-t} \tag{5} $$
we rewrite this as
$$ f_{t}^{'}(t) = e^{-t}t(2 - t) \tag{6}$$
Then we have
$$ t = 0, 2 $$ $$ f_{1}(2) = (2)^{t} e^{-2} = 4 \cdot e^{-2} = \frac{4}{e^{2}} \tag{7} $$
so our maximum for $f(t)$ is given by
$$ f(2) = \frac{4000}{e^{2}} \tag{8} $$