Show that the maximum volume enclosed by a right circular...

Key Concepts

Calculus-based Optimization

Calculus-based optimization involves finding the maximum or minimum of a function by applying differentiation techniques to locate critical points subject to constraints. This method is widely used in problems where one must determine the optimal dimensions of a shape that result in a maximum or minimum measure, such as volume or area.

Inscribed Solids

An inscribed solid is one that is completely contained within another, typically leading to a relationship between their respective dimensions. Understanding these geometric constraints is essential for forming equations that relate the dimensions of the inscribed shape to those of the circumscribing shape.

Volume Ratios

Volume ratios involve comparing the volumes of different solids to assess relative sizes and efficiency. In geometry and optimization, determining the ratio of an optimal inscribed solid's volume to that of the larger solid offers insight into the proportional use of space and the effectiveness of the design.

Transcript

00:01 Show that the maximum volume enclosed by a right circular cylinder inscribed in a sphere is one over root three times the volume of the sphere.

00:09 So we have the sphere shown here and we have the right circular cylinder inscribed inside of it.

00:14 So here we have the center of the sphere and we're going to let the radius of that be r.

00:20 So the volume of that, um, the volume of the cylinder will be given by pi times the radius of the cylinder squared.

00:32 So i'm actually going to redraw this radius a bit here to make it make more sense.

00:44 So there's our circle and let me finish redrawing this back in a bit.

00:55 Okay, so the radius we're going to show goes from the center up to that point where the cylinder intersects the sphere.

01:05 We'll call that r.

01:06 That's the radius of the sphere.

01:08 And then notice how we get a right triangle here.

01:12 We'll let this be x and this be y.

01:15 The radius of that cylinder would just be x and we have this distance y.

01:23 This distance below the center of the sphere would be y.

01:26 So the height of that cylinder is two y.

01:30 So therefore the volume of this cylinder is pi times the radius squared, which would be x squared, times the height, which is two y.

01:40 So i have the volume to be, if we rearrange this a bit, two pi x squared y.

01:46 Now we have a right triangle here that allows us to relate x, y, and r.

01:53 By the pythagorean theorem, x squared plus y squared is equal to r squared.

01:59 This will allow us to take our volume formula for the cylinder and replace x squared with an expression.

02:06 If we solve this formula for x squared, that gives us x squared equals r squared minus y squared.

02:12 And we could substitute that in.

02:14 Volume is two pi.

02:16 And x squared is right here.

02:18 It's r squared minus y squared.

02:20 So we have r squared minus y squared times y.

02:24 We can distribute the two pi and the y to what's in the parentheses.

02:29 That will give us two pi r squared y minus two pi y cubed.

02:37 So we're wanting to maximize this volume formula.

02:41 So that means we need the derivative of this formula, of this equation.

02:44 The derivative of the volume would be the derivative of two pi r squared.

02:49 We need to keep in mind this r is just the radius of the sphere.

02:52 That's a constant.

02:53 That means all of this coefficient with y, two pi r squared, is a constant.

02:58 So the derivative of this constant times y would just be that constant, two pi r squared minus the derivative of two pi y cubed.

03:07 We use the power rule here since we're differentiating with respect to y...

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