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A right circular cylinder is inscribed in a cone with height $ h $ and base radius $ r $. Find the largest possible volume of such a cylinder.

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Section 7

Optimization Problems

Derivatives

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Lectures

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In mathematics, the volume of a solid object is the amount of three-dimensional space enclosed by the boundaries of the object. The volume of a solid of revolution (such as a sphere or cylinder) is calculated by multiplying the area of the base by the height of the solid.

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A review is a form of evaluation, analysis, and judgment of a body of work, such as a book, movie, album, play, software application, video game, or scientific research. Reviews may be used to assess the value of a resource, or to provide a summary of the content of the resource, or to judge the importance of the resource.

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A right circular cylinder …

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Find the dimensions of the…

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Cylinder in a sphere Find …

uh, right Circular cylinder is inscribed in a cone with height H and base radius R find the largest possible volume of such a cylinder. Okay, so here's the picture. Here's the cone, and its radius is our and its height is H and I have ah cylinder inscribed in there, which means the cylinder has to be touching the cone on the base and then at at least two places here, I mean, it has to be touching all the way around. Sorry, because it's a circle. Alright, So here's the question. Is that the one with greatest volume? Or is this one the one with the greatest volume? Or maybe it's a little flat one like this. All right, so first of all, we're trying to find the largest, so we're trying to maximize the volume of the cylinder and volume of a cylinder will be pie. Oh, we can't use H and R. We're gonna have to be a little bit careful here. You know what? Before I do that, let's go ahead and draws a two dimensional picture so we can give everything in name. I'm gonna put it in the X Y coordinate system. Okay? That's the cone. So this length is our This height is age. Okay, here's my cylinder that I'm gonna put in there. So since I put it in there Now I know what to call the radius. Let's call it X. Let's call height. Why? So we're trying to maximize the volume of the cylinder and a spy X squared y hopes. You think because the formulas pi radius squared times height. Okay, Now we have to figure out what the constraint is. Well, the constraint is that this point right here is touching this piece of the, um, cone. Okay, so we've got to figure out if this points name is X y how our X and y related on this line. Okay, where this point is are zero, and this point is zero h. All we gotta do is write the equation of the line. You can see that the slope is down h over our. So why equals? Oh, it's my constraint. Why equals down H over r Times X plus the y intercept, which is h there. All right. So that's her constraint case that forces the cylinder to be inscribed because it's touching. All right, So now let's take our equation, which is V put. V equals Ty X squared y, and we want to take the derivative. So we're gonna use the constraint right here in place of why I'm gonna put this. Can you remember? H and R are constants here. Even though they look weird like that, they're just standing for numbers. All right, Now I'm gonna take the derivative. Okay? So if I take the derivative like this, I'm gonna have to use the product rule. So before I do that, I'm gonna go ahead and multiply everything out here. So I got pie. I'm gonna leave the pie. Outfront. How about that? Minus h over our X cubed plus h X square. Okay. Derivative ISS pie can minus H over our is a constant. So the derivative of execute three X squared plus h is a constant derivative of X squared two X. I want to set that equal to zero. So minus three h over our X squared plus two h x equals zero. I think I will factor M h and X out, so h x. So what? That leaves me is minus three over our X plus two equals zero. So either HX zero or minus three over our X plus two equals zero. Well, we know h is not zero because it's the height of the cone. We know X can't be zero or else we'll have a cylinder of zero volume. So that's not it. So, minus three over our X equals negative too. So X is two times are over three. Okay, now, what was the question? Find the largest possible volume. All right, so remember the volume before I took the derivative waas? All right. Here. Pie. Yeah. Times minus h over our x cubed minus h over our X cubed plus h x squared because it was minus plus h x squared. So pi times minus h over our times two are over three cute plus age times to our over three squared. That's pie minus h over our times. Eight R cubed or 39 27 plus h times for R squared over nine. So that is pie Times 8 20 set minus 8/27 h r squared plus 4/9 h r squared. So get a common denominator Specter of the h r squared out here so minus 8/27 plus three, three, 12/27. So pi H r squared times for over 27. Okay, so that is the maximum volume of a cylinder that is inscribed inside a cone whose radius is our and heights this height is H.

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