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The water level, measured in feet above mean sea level, ofLake Lanier in Georgia, USA, during 2012 can be modeled bythe function$$L(t)=0.01441 t^{3}-0.4177 t^{2}+2.703 t+1060.1$$where $t$ is measured in months since January $1,2012 .$ Estimatewhen the water level was highest during $2012 .$

About 4.1 months after January 1

Calculus 1 / AB

Chapter 3

Applications of Differentiation

Section 1

Maximum and Minimum Values

Derivatives

Applications of the Derivative

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Okay, so for this question were given a function l of tea and arrange tea going from zero up to 12. And we're tasked with finding the maximum value of our function and then reporting the value for T at which that values attained. So to do this, we're gonna want to take the derivative of LFT said it equal to zero and solve for the points of tea that give us those values. The reason we're gonna want to do this is because if we want to find the local maximum values at this function, uh, those are gonna be the spots where the tangent line to our function is equal to zero, which is the equivalent saying that the derivative there is equal to zero. So let's get started. We're gonna be looking for l prime of tea, which will denote the derivative of our function. And all that we have to do here is go term by term, across their polynomial, take the exponents multiplied by the coefficient in front and subtract one off of the power. So for the 1st 1 that's gonna look like 10.1441 times three, which gives us 0.0 four 323 And then we subtract one off of whatever our exponents Waas sore teeth cubed becomes a T squared For the next term, that's gonna be 0.4177 times two, which gives us 0.8 354 and against attracting one off the exponents gives us t to the one which we can just think of as teeth. And then the last one that we have to do this for is the third term of this polynomial, which is 2.703 t. We can think of the exponent of our T here being one so multiplying too point Ah, 703 by one just gives us 2.703 and subtracting one from our exponents on rt will give us zero and t 20 is just one. So we're left with just 2.703 for that last term of our derivative. Now we want to find the spots where this derivative is equal to zero. So we could just set this polynomial equal to zero and solve for t. There's a couple of ways we can do this since we don't have nice whole numbers as our coefficients here, we instead have decimals. I'm going to suggest we use the quadratic equation. Ah, the quadratic equation will tell us to values of T for which dysfunction ah is equal to zero. Uh, the quadratic equation is just negative. B plus or minus the square root of B squared minus four a c all over two a. Us. All right, out quick. What? That would be for this question. Negative B is going to be a positive 0.354 because it is the positive version. Ah, of this number here plus or minus square root of I want to make that even longer. It's gonna be a long square. Ah, four or B squared eso That's that point 83 u 54 the power of to minus four times a, which is 0.0 4323 times. See, which is 2.7 03 Since then, the whole thing will be divided by two times a And for us A. Is that 0.0 for 323 Okay, so if you didn't catch that are a is the term in front of our T squared Barbie is the term in front of our tea and R C Is this term here on its own? So our next up, then, is to just simply plug this into a calculator or something and ah, get some values. If we plug in the 8.354 plus this whole thing, we're gonna end up getting T is equal to 15 0.2. And if we plug in the 0.8354 minus this whole square root thing, we're going to get T is equal to 4.1. So these air two of our potential answers for this question. Uh, but we're not quite done here. Um, the first thing that we can do is notice that 15.2 is actually outside of our range. That we were looking at up here. We said he had to be between zero and 12 so we can right away, just tossed 15.2 out. The next thing that we need to do is make sure that 4.1 is actually a maximum value. And to do this, we're just going to plug it into the function and also checked the endpoints of our function because those were the only two other potential values that could be local maximum Unz. So to do that, uh, let's do this. Not in red, but black. Over here, we can say l a zero equals something. L of four point one equals something, and l love 12 equals something. And so to get these numbers, all that we have to do is plug these 0.4 point one and 12 into our function here, the original function, and see what numbers we get when we plug in zero. All the terms dropped way. Except that last term. So we can do that. One with the out, even plugging it into a calculator. And we're gonna get 1060.1. If we plug the 4.1 in, we end up getting 1000 65 0.2. And if we plug in 12 we end up getting 1057.1. Whoops. Point three 0.3. Okay, So of these, the largest value is 1065.2, and that value was attained. Uh, T is equal to 4.1

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