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

## $$4 \pi r^{3} /(3 \sqrt{3})$$

Derivatives

Differentiation

Volume

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### Video Transcript

So what we have here is a side view of a cylinder inscribed in a sphere of radius R So we'll have our sphere here, which looks like a circle. And then we have This is a cylinder because from a side view of cylinder just looks like a rectangle. So this is point A we'll call this point B point B. This will be point D and this will be point c. We know that point is right here in the center. We have our radius r and then we have this length right here, which will be big are this is a church over to? We'll call this point f and this point e and H is this whole long side right here. So we want Thio take thes and find the volume. We already know that the volume of the cylinder of a normal cylinder is pi r squared h Then we need another equation to relate this with the other variable. And that's going to be on the Pythagorean theorem we see that are squared plus h over. Um let's see we have If we solve for r squared, we'll get that h squared r squared is equal to R squared minus h squared over four. And that's the A squared plus B squared equals C squared. We have a squared plus b squared equals C squared. And then we just subtracted one of the components of the other side. So now we can sub substitute for r squared into our volume equation, so we'll put this right here. This is gonna go right in there. So now what we have is that the volume is equal to pi R squared minus h squared over four h. So now what we have is that this is equal to pi times R squared H minus h cubed over four. So now we're gonna want to differentiate this so v prime. Well, give us, um, pie times are squared minus 3/4 each square. When we solve for it at zero, we end up getting that H is equal to the square root of four thirds R squared. So it's just gonna be equal to two over rad three are. Then we can plug this value into R V double prime. We know that V double prime is going to equal pie times negative three over to each So when we plug in the value into original the we'll get the V equals pi. Times R squared two over route three are minus chew over. Route three are cute over four, and this is going back to our original equation right here. Then we simplify this further and we get that V is equal to 4/3. Route three. Hi r cubed. So that is going to be the volume that we find, um, for this particular cylinder.

California Baptist University

Derivatives

Differentiation

Volume

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