💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

Like

Report

Numerade Educator

Like

Report

Problem 25 Medium Difficulty

Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius $ r $.

Answer

2$r^{2}$

Discussion

You must be signed in to discuss.

Video Transcript

so, whereas to find dimensions of the rectangle of largest area that commune inscribed in a circle of radius R. So we have a circle of radius r so and we have a rectangle inside of this of maximum area. I love for you. That small hole is our way. Just so there. That's damn I love wearing its dominance Now the screen is mhm, mhm. He's like a little more garages. Current events. We have the area. Such a rectangle if we call the length of one side X and the other side y is going to be X times y now notice that we can also draw a point from the center of the circle to one of the corners of the rectangle, which has a length between them. A segment of length are so by Pythagorean theorem we have that half of the length of one side X over two squared, plus half of the length of the other side. Why over two squared is equal to Hi pa tennis R squared. Yeah, this way we have a relationship between X and Y. Oh, now to help us notice that function area is always greater than or equal to zero, right? Really? Yes. Well, we can't quite do that, But notice that both x and Y have to be greater than zero. So we can solve for why, in this equation we have Why over two squared equals R squared minus X over two squared so that why over is equal to two times the positive squared of R squared minus x over two squared. Yes. And therefore area is a function of X is X times two times the square root of R squared minus x over two squared. Great. If they started. Yes, by the website doesn't seem list. Yeah, so now you want to find a maximum value of a I notice that X and Y are both also less than our Uh huh. So do this. I'm going to find the derivative a prime of X. This is going to be well before we even do that. Actually notice that a is, of course, always greater than or equal to zero on zero to our yeah, and at a this is they therefore has a maximum When the function F, which is a squared, is maximized now F of X, this is equal to four x squared times R Squared minus x over two squared now find the max move f f Prime of X is buy the product rule. This is two times 48 x times R Squared minus x over two squared plus four x squared times. The derivative The second factor. This is negative two x over two times one half. We're going to set this equal to zero to find the critical values. So we have eight x r squared minus eight or four is too X cubed. Yeah, plus, let's see. This is negative. Four over two is negative two, and this is again execute equal zero. So we have positive. So there should be negative to execute course so positive. For X cubed minus eight R Squared X equals zero we can solve by factoring we get four x times X squared minus actual Jamaica two R squared equals zero What not a single white here. This can also be factored as four x times X minus route to our times X plus route to our so the solutions and are critical values are X equals zero and X equals plus or minus route to our it's they were now determine for which of these critical values Yeah, f is maximized subject to X between zero and are well, first of all, plus or minus were to our is already not an option word. I'm sorry. Made a steak here extra actually be between zero and two are so while negative route to our is not necessarily good as a tribute to our could work. So really, just looking at positive Group two are and zero now, the second derivative F double prime of X. Hi, this is mhm uh, eight R squared minus 12 x squared, which we plug in zero after the prime of zero is eight r squared, which of course, is positive. And after the prime of route to our is eight r squared minus 12 times two r squared, which is negative 16 r squared, which is less than zero now because second derivative of f at zero yes is positive. It follows that f has a local minimum at X equals zero and a local maximum at X equals our route to because, well, now we need to compare the values of X equals our route to and to our So we have f of to our mhm Listen, So this is on 0 to 2 are f of two rs four times for R squared. Yeah, times are squared minus yes, R squared, which is zero while f of yeah are too. This is four times two r squared times R squared minus route to over two squared, which is one half r squared. We're like this is eight r squared times one half r squared, which is for art of the fourth, which is positive. Yeah. Therefore it follows the f has and absolute maximum on the interval. 0 to 2 are at X equals route to our now to find the other dimensions of the rectangle We know that why is going to be mhm two times the square root of r squared minus x Super two squared. This is two times the square root of R squared minus. There too, are over to squared, which is equal to two times the square root of R squared minus one half R squared, which is to over root two times are which is the same as route to our. Therefore, the dimensions are X equals are too and why so equals R two