Find the height h, radius r, and volume of a right circular cy...

Question

Find the height h, radius r, and volume of a right circular cylinder with maximum volume that is inscribed in a sphere of radius R.

Accepted Answer

Welcome back, everyone. A cone is inscribed in a sphere of radio 6 units. Find the height of the cone that maximizes its volume. A says it's 6 units, B 7 units, C 8 units, and the D 12 units. Now to find the height of the cone that maximizes its volume, let's try to visualize what's going on here, OK? So let me come over to the right here and let me draw the cross section of our sphere with our cone inscribed in our, in inside our sphere, OK? And now we're assuming here that the base of our cone is parallel to the horizontal plane and the The the vertex of the corn is at the bottom of the sphere, OK. No, when we do that. We know that we have the height of our coin here, so let's call that height H. OK. And now we also know if we were to treat the put the center at the origin of the sphere. Then from the origin to one edge of our cone would be the length of our radius. Let's call that radius R, OK. From the midpoint of our Of our cone to the edge is going to be the radius of the cone. So let's call that or let's give that value a common R. And then if we look at our diagram here, if the entire vertical length is H, and if we look at the length from the center of our sphere to the vertex, which is going to be the radius 6 units, then that means from the base of the cone to the center of our sphere is going to be the difference between 8 and 6. In other words, H minus 6. No, in that case here then. If we're going to find a height that maximizes the volume, we'll need to find some way to express our volume purely in terms of the height of the coin. Now let's first see if we can try to do that. Let's think about our cone here. If you take a good look, you notice that a right triangle is formed between the radius, the base of the cone, and the height, or sorry, the length from the base of the cone to the center of the sphere. So we can relate these three quantities using the Pythagorean theorem. And by the Pythagorean theorem. It basically tells us, OK, that. The radius, sorry, the radius of our cone are squad will be the difference between 6. Which is the radius of our sphere, so that's 6 squared and over length H minus 6. So 62 minus H minus 6 squared. I know we could rewrite that as 36 minus H minus 6 squared. Now, since we have our radius R in terms of our height H, we could go ahead and solve for R but let's ask ourselves, what do we know about the volume of our corn? Well, we know that the volume of a corn V. is equal to a third of pi R2 H. OK. So since we have an expression here for R2d, we can substitute it into our formula for the volume to express our volume purely in terms of the height of a coin. So no, it's going to be 1/3 of pi. Multiplied by 36 minus H minus 6 squared, OK. Multiplied by H. If we expand H minus 6 inside our bracket here. It's going to be minus, well, remember, now we're subtracting what we've expanded. So it's minus 82, it leaves us with 82 minus -12H. So that's gonna be plus 12H minus 36, so that's gonna be minus 36, OK? And we're still multiplying all of that by H. Now 36 minus 36 equals 0. So we're going to be multiplying H by negative H2 and 12 H. and when we do that, then we should get our volume to be a third of pi multiplied by 12 H2 minus h cubed. No, we can go ahead and maximize our volume with respect to the height to find a maximum height. And to, to maximize it, we're going to differentiate it, set it to zero, solve for age, and then test if that critical point is a maximum, OK? So maximizing. The of age. To find the maximum height. Remember we're going to test it afterwards, OK. Then no, we're going to find a derivative of VFH. So when we differentiate V with respect to H, then we should get it to be a third of pi multiplied by 24H minus 3H2. And this is what we're going to set to 0. If we multiply through by 1/3. Then we're left with pi multiplied by 8 H minus 8 squared equal to 0. And if we divide through by pi and factor of H, then we're going to have H multiplied by 8 minus H equals 0. If that's the case, then either H equals 0 or H is going to be equal to 8. But now if we think about it, if our height is 0, then that means our volume of the cone will be zero, but that's not possible. So that means that H equals 0 is not a valid solution. Therefore, H equals 8 units. Is going to be our possible maximum. It's a critical point, but remember we don't know if it's a maximum yet. Now that we have that value, let's test if it's a maximum, yeah, that's what we'll need to do next. So let's test if H is a maximum. Now we can know if H is a maximum by trying to understand the behavior of the first derivative before and after H equals 8. If the first derivative is positive before H equals 8 and negative after H equals 8, that means the volume is a maximum. So let's go ahead and test that. Now, for H, less than 8, let's say that H equals 7.9. Then our first derivative. OK, let me just clean that up. Our first derivative of V of H is gonna be of 7.9, is going to be equal to pi multiplied by 8 multiplied by 7.9 minus 7.9 squared. And this is approximately equal to 2.48. So before H is equal to 8, our value is positive. Now what about after? Well, for H greater than 8, let's let H be equal to say, 8.1. No, our first derivative. is going to be equal to pi multiplied by 8 multiplied by 8.1 minus 8.1 squared. And this is going to be approximately equal to -2.54. Now what do we notice? Our sign went from positive to negative. So in that case, that tells us then. That it is a maximum. Thus, this confirms that our volume is maximum at H equals 8 units. If we look back at our answer choices, that means that C is the correct answer. Thanks for watching everyone. I hope this video helped.

Find the height h, radius r, and volume of a right circular cylinder with maximum volume that is inscribed in a sphere of radius R.

Key Concepts

Right Circular Cylinder

Volume of a Cylinder

Inscribed Shapes

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