What happens to $e^{-0.2t}$ as $t$ becomes very large? How does that affect the value of $M(t)$?

For a logistic curve, the graph rises and then flattens out near a horizontal line. In a real-world situation like a growing culture, what does that upper horizontal value represent?

In Desmos, type M(t) = 250/(1+7e^(-0.2t)) and make sure the viewing window shows $t \ge 0$ and a reasonable range of $M(t)$ values (for example, $0 \le t \le 50$ and $0 \le M \le 260$).

Click on the graph at $t = 0$ or add the point (0, M(0)) by typing it. Note the $y$-value there; this shows you the initial mass and lets you see that it is not 250.

Pan or zoom right along the $t$-axis (increasing $t$) and watch how the graph behaves. Notice the horizontal line that the graph gets closer and closer to but does not cross. Read off the $y$-value of that horizontal level and think about what that value means for the culture’s mass over time.

The given function is

A logistic function often has the form

where $L$ is a constant that the function approaches as $t$ becomes very large (the horizontal asymptote). Here, $L = 250$.

To see if 250 is the initial mass, plug in $t = 0$:

So the initial mass is $31.25$ milligrams, not $250$, which eliminates any interpretation that treats 250 as the starting value.

As time increases, the exponent $-0.2t$ becomes a large negative number, so $e^{-0.2t}$ gets very close to $0$.

That makes the denominator $1 + 7e^{-0.2t}$ get very close to $1$, so the whole fraction $M(t)$ gets very close to $250$ for large $t$. This means $M(t)$ approaches 250 but does not exceed it.

In a biological context, when a logistic model levels off at a certain value, that value represents the culture’s limiting size under the given conditions.

Since $M(t)$ approaches $250$ milligrams as $t$ increases, 250 represents the maximum possible mass the culture can reach according to the model. Therefore, the correct choice is D) The maximum possible mass the culture can reach according to the model.