Problem

Find the dimensions of the rectangle of largest a…

View

Problem 26 Medium Difficulty

Find the area of the largest rectangle that can be inscribed in the ellipse $ x^2/a^2 + y^2/b^2 = 1 $.

Name: find the area of the largest rectangle that can be inscribed in the ellipse x2a2 y2b2 1
Uploaded: 2021-04-02
Duration: 9 min 39 s
Channel: Numerade
Description: Find the area of the largest rectangle that can be inscribed in the ellipse $ x^2/a^2 + y^2/b^2 = 1 $.

Answer

2$a b$

Grace H.

Heather Z.

Oregon State University

Samuel H.

University of Nottingham

Joseph L.

Boston College

Watch More Solved Questions in Chapter 4

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 73

Problem 74

Problem 75

Problem 76

Problem 77

Problem 78

Problem 79

Problem 80

Problem 81

Problem 82

Video Transcript

there has to find the area of the largest rectangle that can be described in the Ellipse X squared three squared plus y squared over B squared equals one. Mhm. Mm. A lot of these hotels lately. And so imagine we have the sketch of this ellipse. Don't know so much. It was in France. We have a major axis of length in this case to a That's it. And be okay. Minor axis of length to V Now inside, the Ellipse will draw a rectangle like you do. Like Great. So now the width of this will be well, we'll call this. Uh I have a question. Um, you see. Ah, Is it possible? Yes, two times X and the height is two times y by symmetry. Yeah. So the area of the largest rectangle. Well, the area in general of the rectangle is two x times two. Why is four x y? We're both x and y are positive part Reynolds gave and solving for why? Hey, Bert, we have why squared is equal to B squared times one minus x square brace squared so that why is equal to the positive square root. This is just be times the square root of one minus X squared over a squared. This is the same as the over eight times the square root of a squared minus X squared. Plugging this into a We have a as a function of X is four x times do you over eight times the square root of a squared minus X squared. Now we know that that's it. Our area A is in fact, always positive for X greater than zero and X also has to be less than a so So it follows that the function f which is starting. Yeah. Sorry. So the function a is very so maximized. Yeah. I mean, that's like the Tim Allen when the function F, which is a squared, is maximized. So as a function of X, this is a squared of X, which is C 16 b squared over a squared X squared times a squared minus X squared. Now, to find the area of, well, the largest value of F the absolute max, we're gonna have to find the derivative of F prime of X. This is 16 b squared over a squared times two x times a squared, minus X squared plus X squared times. Negative two x, and we want to find the critical values. So we set X equal to zero. No itself. Really. Now this is only equal to zero. When to a squared X minus four X cubed equals zero or two X times two X squared minus a squared equals zero. Hey, it's me, Burt Reynolds work. It's working Herb Reynolds here. Damn, obviously. And so, in fact, er, this is two x times route to X minus a time is Route two X plus a equals zero first. But when the Mark Walberg and Ellen de Generous hilariously create convert so we have critical values X equals zero a over root two plus or minus Eva roof to I should say, however, we're only considering values of X between zero and a. So I'm just considering X equals zero a of a route to we saw I would say, Let's find the secondary, the double Planet X. Guess you asked the gym. So this is mhm 16 b squared over a squared yeah, times to a squared a minus 12 x squared. Now it's your Syria and we have the F prime of zero. Well, this is 16 B squared time is, too, which is clearly greater than zero. So it follows that F has a local minimum. I went at X equals zero. Let's look at our other critical value. It's inevitable. Prime of a over root two. Well, this is 16 B squared over a squared and two way squared minus 12 times a squared over two with minus six a squared. Just negative, right. Four times 16 B squared, which is less than zero. Therefore, by the second derivative test F has a local maximum. Speak at X equals over to 17. Now, let's compare the value of F A over two and the other end point PlayStation. Mm, Yeah, X equals a. I have a favor too. This is 16 b squared over a squared times a squared over two. Yeah, you got a real threat. Your career times a squared minus a squared over two, Which is right. Yeah, four a squared B squared, mhm. And on the other hand, f of a. This is just zero. So So it follows that f has in fact, an absolute maximum at X equals mhm over to and therefore it follows that our maximum area A is A f. Avery, too, which we plug this in. This is four times Eva route to time to be over a times the square root of a squared minus a squared over two, which is okay yourself something to a B talk. This is the area of the largest rectangle inscribed in the Ellipse.

Chris T.

Ohio State University

Related Courses

Calculus 1 / AB

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems