# Video: Using the Logistic Model for Population Growth

An aquarium contains fish with a carrying capacity of 1,200 and a growth rate of 8%. If the initial population of the fish is 400, what is the population of the fish at any given time?

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### Video Transcript

An aquarium contains fish with a carrying capacity of 1200 and a growth rate of 8 percent. If the initial population of the fish is 400, what is the population of the fish at any given time?

In this question, we need to consider models for population growth. But one of the key features is that this aquarium has a carrying capacity of 1200, which means that this population of fish cannot grow in an unrestricted way. There is a maximum population of fish that can be supported in this aquarium. Now, this tells us that instead of using the simple model for population growth, we instead need to use the logistic model for population growth, which incorporates this feature that there is a maximum population that can be supported.

This logistic model, which is the solution to the logistic differential equation for population growth, has the following general form: the population at time 𝑡 is given by 𝐿 over one plus 𝐴𝑒 to the power of negative 𝐾𝑡. Where 𝐿 represents the carrying capacity of the population, 𝐾 is its growth rate, and 𝐴 is equal to 𝐿 minus 𝑃 nought over 𝑃 nought, where 𝑃 nought is the initial population. We can determine several of these values from the information given in the question. We’re told that the carrying capacity — so that’s 𝐿 — is 1200. We’re also told that the growth rate is 8 percent. So, as a decimal, that’s 0.08. And we’re told that the initial population of the fish is 400. So, that’s our value for 𝑃 nought.

We can, therefore, calculate the value of the constant 𝐴. It’s 𝐿 minus 𝑃 nought over 𝑃 nought, so 1200 minus 400 over 400. That’s 800 over 400 which is equal to two. To find the population of fish at any given time then — so that’s an expression for the population 𝑃 in terms of 𝑡 — we just need to substitute the values of 𝐿, 𝐴, and 𝐾 into the logistic model. Doing so gives that the population at time 𝑡 is equal to 1200 over one plus two 𝑒 to the power of negative 0.08𝑡. Now, this would be a perfectly acceptable form in which we leave our answer. But we may prefer to deal with positive exponents for our exponential rather than negative.

We can achieve this by multiplying both the numerator and denominator of our fraction by 𝑒 to the power of positive 0.08𝑡. And as we’ve multiplied both the numerator and denominator by the same, we’ve created an equivalent expression. Doing so gives 1200𝑒 to the power of 0.08𝑡 in the numerator. And in the denominator, we have one multiplied by 𝑒 to 0.08𝑡. And then, we have two 𝑒 to the negative 0.08𝑡 multiplied by 𝑒 to the power of 0.08𝑡. So, we recall here that we add the exponents together, giving two 𝑒 to the power of zero. But of course, 𝑒 to the power of zero is just one. So, this simplifies to two.

So, by using the logistic model for population growth, we found that the population of fish at any given time in this aquarium, which has a carrying capacity of 1200, a growth rate of 8 percent, and an initial population of 400, is given by 𝑃 of 𝑡 equals 1200𝑒 to the power 0.08𝑡 over two plus 𝑒 to the power of 0.08𝑡. We can find the population any given time by substituting the relevant value of 𝑡.