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

2$a b$

05:03

Wen Z.

01:39

Amrita B.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems

Derivatives

Differentiation

Volume

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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.

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