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The water level, measured in feet above mean sea level, of Lake Lanier in Georgia, USA, during 2012 can be modeled by the function

$$ L(t) = 0.01441t^3 - 0.4177t^3 + 2.703t + 1060.1 $$

where $t$ is measured in months since January 1, 2012. Estimate when the water level was highest during 2012.

$L(t)=0.01441 t^{3}-0.4177 t^{2}+2.703 t+1060.1 \Rightarrow L^{\prime}(t)=0.04323 t^{2}-0.8354 t+2.703 .$ Use the quadratic formula

to solve $L^{\prime}(t)=0 . \quad t=\frac{0.8354 \pm \sqrt{(0.8354)^{2}-4(0.04323)(2.703)}}{2(0.04323)} \approx 4.1$ or $15.2 .$ For $0 \leq t \leq 12,$ we have

$L(0)=1060.1, L(4.1) \approx 1065.2,$ and $L(12) \approx 1057.3 .$ Thus, the water level was highest during 2012 about 4.1 months

after January 1.

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So if we want to try to find the maximum, um, water level. Given this equation here, the first thing we have to remember is so since we're doing this on Lee for a year and our time is supposed to being months from January 1st we're really working on this interval, 0 to 12 were essentially applying the extreme value there to this. So what will need to check outside of just Oh, whatever this derivative is, we need to check l of zero as well as l of 12. So let's actually go ahead and do those first S o l of zero. If we were to plug that in, that's actually just gonna be 1060.1. And then when we plug in 12 into this, we get something around 1500 or 1057.3. Yeah. So these were going to be our in points. And now we just want to see Okay, Is there some local max that lives within this interval? Now? So for a simple dio take the derivative, So l prime of teeth, uh, then to take all these derivatives, we're just going to use power rule, so we would have 0.4323 t squared power rule again. So that would be minus 0.8354 and then plus 2.703 Right now, remember, we want to set this equal to zero, and for us to set this equal to zero, we could just use the quadratic formula to solve it. So that would tell us he is going to be equal to so negative be so 0.8354 plus reminds the square root of B squared eso 0.8354 crash. I'll just write this all out first, and then we could do the calculation afterwards. So this squared e I guess it doesn't really matter if it's negative sense were squaring it and then minus for a C. So I shall just go ahead of multiple that together. So for times 0.4323 times 2.703 And that's going to give 0.467 or, uh, actually just kind of rounded there. So I'll kind of start approximating this, um and then to a so that would be two times 20.1441 So 0.2882 Okay on. Now, if we were to go ahead and plug everything and we end up with two different solutions. So we either get one or 10.1 or about 15.2. And now, since we have those well, first notice that 15.2 does not live in our interval of 0 12, so we can actually throw this one out, and we only want to check or 4.1 here. So now we'll do 4.1, and we'll come back up here and plug that in. So we'll do l of 4.1, and this is going to be approximately, um, 1065.2. All right, so now we get to compare these three so are in points as well as our critical point. And so, the largest out of all of these, it looks like it is this one right here. So the largest water level we will reach on what is three units for the supposed to be, um, feet. So this is just going to be 1065.2 ft above sea level. And so that will be the maximum. Um oh, actually, but it wants us to find what time? Um, it waas so it occurs at this. But it also was 4.1 months after January 1st. Oh, which would be so January, February, March, April, May. So it's around May or so that this would happen so around May. But so this is when we would have when it occurs, though.

University of North Texas