Covering a marble Imagine a flat-bottomed cylindrical pot with...

Question

Covering a marble Imagine a flat-bottomed cylindrical pot with a circular cross section of radius 4. A marble with radius 0 < r < 4 is placed in the bottom of the pot. What is the radius of the marble that requires the most water to cover it completely?

Accepted Answer

Welcome back, everyone. In this problem, an empty cylindrical container with a circular base of radius 6 is standing upright. A solid sphere with a radius R that's greater than 0 but less than 6 is placed at the bottom. Water is gradually added until the sphere is fully submerged. Determine the radius of the sphere that requires the maximum volume of water to submerge it completely. A says it's 3 Route 3 units, B 2 route 2 units, C Route 2 units, and D 3 route 2 units. Now, here, if we're going to figure out the radius of the sphere that requires the maximum volume of water, let's draw a quick diagram to help us understand exactly what we're looking for, OK? So we're told that we have an empty cylindrical container, probably looks something like this, OK. And it has a circular base with a radius of 6, and it's standing upright. OK, so it has a radius R. Now, a solid sphere is placed at the bottom, so we can draw our sphere here. Let me try to, my sphere is not gonna be perfect, but it's probably gonna look something like that, OK? And then we're told that the water is added until the sphere is fully submerged. So we can imagine here then that the top of the sphere is at the top of the watermark and the bottom of the sphere is at the bottom of the cylinder. Now in this case we know a sphere has a radius of R, OK, and that means then that the height of the water. It is going to be 2 hours because the height of the water will be the diameter of the sphere. Now, how are we going to maximize the volume of water? What do we know about the volume of water? No, the volume of water. OK, is going to be the difference between the volume of our cylinder, OK? And the volume of our sphere. What do we know about both of those volumes? Well, the volume of our cylinder, OK, is going to be pi R2 H. So that's pi multiplied by 62 multiplied by H, OK? And we know. That H equals 2 R. OK, so we could rewrite this as. Pi multiplied by 6 grade, so that's 36 pi multiplied by 2R. And no, when we simplify that, it's gonna be 72 pie R. That's the volume of our cylinder. Next, we also know that the volume of our sphere, OK, is going to be equal to 4/3 of pi are cubed. So now what this tells us then. Is that our volume of water as a function of our radius R is going to be equal to 72R minus 4/3 of piR cubed. So to maximize the volume of water, we can differentiate the volume of water with respect to R and set it to 0, OK? So let's go ahead and do that. Let me put that in red, OK? So in this next step, we are maximizing our volume of water. Now, to maximize it, that means we're going to differentiate it and set it to 0, and now the derivative of our volume function. Is going to be equal to 72 minus 4 pi R squad because we can differentiate both of these using the poor rule of differentiation. Now setting our first derivatives to 0. OK. Then we're going to have 72 minus 4 pi R2 equal to 0. If we divide through by pi, OK, and subtract and add 4 R squad to both sides, OK? Then we're going to get 4 R squared. To be equal to 72. And no, that means we'll get, if we divide through by 4, R squared is 18 and R, oops, let me fix that. R is going to be equal to the square root of 18. Why is my 8? Why does my it keep doing that? My apologies. That's gonna be Route 9 multiplied by route 2, which is gonna be 3 route 2. So I is 3 route 2 units. Now that we have our critical value. This is the critical, a critical value for our function, but we don't know if it's a maximum or minimum. So let's test if it corresponds to the maximum so we can maximize our value, OK. So confirming whether R is a maximum. We can use that using the 2nd derivative test, and the second derivative test basically tells us that if the second derivative is negative, then the function is concave done and thus our value is a maximum. But if it's positive, the function is concave up, which means that it's a minimum. So in this case, the second derivative for the volume of our water with respect to R will be the derivative with respect to R of 72 minus 4 pi R squared and that is gonna be equal to -8 R. Our second derivative is negative, OK? So since it's negative, then that means. Our secondary or function, sorry. That means, or sorry, our value R, which is 3 root 2. Is concave done? And thus and corresponds to a maximum. I responds to a maximum. So what does this tell us? Well, no, that means then. That the value of R, OK, the value of R, the radius that the radius of the sphere that requires the maximum value of water to submerge it completely is 3 route 2 units. If we go back to our answer choices, that means D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Covering a marble Imagine a flat-bottomed cylindrical pot with a circular cross section of radius 4. A marble with radius 0 < r < 4 is placed in the bottom of the pot. What is the radius of the marble that requires the most water to cover it completely?

Key Concepts

Volume of a Cylinder

Volume of a Sphere

Optimization

Watch next