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The population of a culture of bacteria is modeled by the logistic equation:

$P(t)\;=\;\frac{14,250}{1+29\cdot e^{-0.62t}}$

To the nearest tenth, how many days will it take the culture to reach 75% of its carrying capacity?

What is the carrying capacity?

What is the initial population for the model?

$P(0)\;=\;\frac{14,250}{1+29\cdot e^{-0.62\cdot0}}\\P(0)\;=\;\frac{14,250}{30}=475\\P(0)\;=\;P_0\;=\;475\\$

Now, we got $P_0$.

75% is $\frac{3}{4}$.

So,

$P(t)\;=\;P_0e^{k\cdot t}\\\frac34\;=\;475\cdot e^{k\cdot t}$

I'm stuck at finding $k$.

Not sure I'm approaching in the right way.

Nay Sie
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    Correct me if I'm wrong, but isn't the 'carrying capacity' the maximum attainable value of $P(x)$, i.e. $14250$? – Vishu Jul 17 '20 at 14:54
  • @Tavish, according to logistic equation, yes. No if it is a different kind of logistic equation. I don't know about it. – Nay Sie Jul 17 '20 at 14:59
  • What do you mean by a different logistic equation? – Vishu Jul 17 '20 at 15:01

1 Answers1

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The carrying capacity is the maximum value of $P(x)$. Maximising it requires minimising the denominator, which is $$1+29e^{-0.62t} $$ and it is achieved when $t\to \infty$, and the denominator goes to $1$. The carrying capacity hence is $$14250$$ Now, you need to find $t$ for which $P(t)=75\%$ of $14250$, i.e. $$\frac 34 \times 14250 =\frac{14250}{1+29e^{-0.62 t}} \\ \implies29e^{-0.62t}=\frac 13 \\ \implies e^{0.62t}=87 \\ \implies t=\frac{\ln 87}{0.62}\approx 7.2$$

Vishu
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