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A right circular cylinder is inscribed in a sphere of radius $ r $. Find the largest possible surface area of such a cylinder.

Maximum surface area $=\pi r^{2}(1+\sqrt{5})$

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Missouri State University

Oregon State University

University of Michigan - Ann Arbor

does pull the right circular cylinder is inscribed in a sphere of radius R and we're asked to find the largest possible surface area of such a sphere. Morning. So I'll draw a diagram, please Show me your yes, We call those burgers. So this is the sphere the right circular cylinder out here, our son like now, the big sphere as a radius of our notice that this is the distance from the center of the sphere to the corners here of the cylinder you're good at. That's how. Now let's call the height of this cylinder H and the radius of the cylinder. We'll call this X. Mm. And by Pythagorean theorem, we have that h over two is equal to the square root of little R squared minus X squared. And it's a now the surface area of a cylinder. I'll call it s a or maybe just s. This is going to be so the areas of the, uh, the side this is Yes. Hi. Yes, two pi x times H plus the area of the circles on the ends, which are two pi x squared. We can write this as a function of X by substituting for H This is equal to SFX is two pi x times So two times the square root of R squared minus X squared plus two pi X squared. Now, to find the largest possible surface area, we're going to use derivatives. Swell, differentiate So as prime of X, it's four pi times the square root of R squared minus X squared plus four pi x times negative x over the square root of R squared minus X squared plus four pi X We'll solve the equation as planet X equals zero to find the critical points. Yeah, elections. Yeah. So we have four pi times. Read it like he looked at his 1500 guys, then set second group sky. Yeah, Okay, Mhm. This is four pi times R squared minus X squared minus four pi times x squared all over the square root of R squared minus X squared plus four pi X equals zero. The solving for X. Yeah, well, I have X equals on one side. I have the radical function, My son Negative. Close to R squared minus two x squared over square root of R squared minus X squared was gay. Mhm. Now that's all that's all square both sides. Uh huh. We get X squared equals. See, this is R squared minus two x squared, squared all over r squared minus X squared across multiplying. I get X squared times R squared minus X to the fourth and this is equal to arch. The fourth minus two times two is four r squared, X squared plus four x to the fourth. Oh, so we get zero equals See are to the fourth minus five r squared X squared plus five x to the fourth. We can write this as a quadratic equation. You know, five times X squared, squared minus five R squared times X squared plus our to the fourth equals zero. Just say the what Using quadratic formula can find X squared. So X squared is negative. B which is positive. Five R squared plus or minus the square root of the square, which is 25 out of the fourth minus four times A, which is five minus 20 times are to the fourth all over two times five, which is 10. Yes, this is five r squared, plus or minus the square root of five are to the fourth over 10 for the home artist. This is equal to five plus or minus. The square root of five times are squared over 10. However, we actually have an extraneous solution. If you try to plug in five minus Route five X squared over 10 to the original equation or to this equation here does not work. So we just have The X squared is equal to five, plus the square to five times are squared over 10. Therefore, we find that X is going to be and because X is a positive length, this is the positive square root five plus Route five R squared over 10, which is equal to our times. The square root of five plus Route five over 10 Now also suit this value of X into our equation for the surface area. Our Max surface area is s of the square are times the square root of five plus route 5/10. And there's a lot of algebra to do here, but you should get this reduces to pi R squared times one plus route five. See if you can find this on your own. Therefore, yeah, this is the largest possible surface area of a cylinder inscribed in a spirit of radius R yeah,