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Problem

On May $7,1992,$ the space shuttle Endeavour was …

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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 53 Problem 54 Problem 55 Problem 56 Problem 57 Problem 58 Problem 59 Problem 60 Problem 61 Problem 62 Problem 63 Problem 64 Problem 65 Problem 66 Problem 67 Problem 68 Problem 69 Problem 70 Problem 71 Problem 72

Problem 65 Hard Difficulty

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

Answer

About 4.1 months after January 1

Related Courses

Calculus 1 / AB

Calculus

Chapter 3

Applications of Differentiation

Section 1

Maximum and Minimum Values

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Applications of the Derivative

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Watch More Solved Questions in Chapter 3

Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
Problem 22
Problem 23
Problem 24
Problem 25
Problem 26
Problem 27
Problem 28
Problem 29
Problem 30
Problem 31
Problem 32
Problem 33
Problem 34
Problem 35
Problem 36
Problem 37
Problem 38
Problem 39
Problem 40
Problem 41
Problem 42
Problem 43
Problem 44
Problem 45
Problem 46
Problem 47
Problem 48
Problem 49
Problem 50
Problem 51
Problem 52
Problem 53
Problem 54
Problem 55
Problem 56
Problem 57
Problem 58
Problem 59
Problem 60
Problem 61
Problem 62
Problem 63
Problem 64
Problem 65
Problem 66
Problem 67
Problem 68
Problem 69
Problem 70
Problem 71
Problem 72

Video Transcript

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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