6d After how many days is the percent of the population infected a maximum

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SUMMARY

The discussion focuses on the infection rate of a disease in College Station, Texas, modeled by the function \( p(t) = 10e^{-t/8} \) for \( 0 \le t \le 40 \). The maximum infection occurs at 8 days, with a peak infection rate of 30%. The derivative \( p'(t) = -\frac{5}{4}e^{-t/8}(t-8) \) confirms that the maximum infection point is derived from setting the derivative to zero, simplifying the analysis of the infection curve.

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karush
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6d
Anotherdisease hit the the chronically ill town of College Station, Texas.
This time the percent of the population infected by the disease $t$ days after it hits town is approximateled by
$$p(t)=10e^{-t/8},0 \le t \le 40$$
a. After how many days is the percent of the population infected a maximum?
$\color{red}{8 \, days}$
b.What is the maximum percent of the population infected?}
$\color{red}{30 \%}$

red is mine
ok got this only by looking a desmos graph
$p'(t)=10{e}^{-\frac{t}{8}}-\dfrac{5t{e}^{-\frac{t}{8}}}{4}$

thot setting $p'$ to zero would answer both question but could do the calculation
 
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Okay, judging by your derivative, we are actually given:

$$p(t)=10te^{-\frac{t}{8}}$$

And so:

$$p'(t)=-\frac{5}{4}e^{-\frac{t}{8}}(t-8)$$

Do you see how that is a factorization of the derivative you gave?
 
yeah that was much easier
 

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