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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.

## Maximum volume of cylinder is $\frac{4 \pi r^{2} h}{27}$

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### Video Transcript

suppose you want to find the largest possible volume of a right circular cylinder inscribed in a cone with radius or and height H. Now, since the radius and height of the cylinder are not defined, Then we let X be the radius and why be the height of the cylinder. And for better understanding, we will draw an illustration for this problem. Now, for optimization problems, we always first find the objective function. This function is the one which is being maximized or minimized in this problem. And so it should be the volume of the right circular cylinder, which is V equal to pi times the square of the radius, that is X squared times the height Y. And then from here we want to write the objective function in terms of one variable only. So either X or Y. So to do that, we need to find a relationship between X and Y. And we will use the given illustration here. So, and here we have the height of the colon, just H. And this radius are here which forms a large strangle. And over this side we see a smaller triangle with height equal to h minus Y. And basically just equal to X. And so by similar triangles, we have the ratio of the height and the base which is H over our that's equal to H minus Y over X. And then from here we want to solve for Y in terms of X. So we have a church over our times X. This is equal to H minus Y. Or we have y equal to h minus H. over our X. And so V is equal to pi times X squared times Y. Which is H minus H over our X. And then from here we only distribute the X squared. So we get We were just five times H X squared minus H over our X cube. Yeah. Next we want to find the critical points of the function that means you find the derivative of V. And he said it's zero and then solve for X hole, the derivative of V. This is equal to pi times we have H. Times two X minus. We have H over our times three X squared. And then from here we factor at the G. C. F. That's pi times H. X times you have two minus three over our X. And then in here we want to set this to zero and Sulfur X. So we have zero equal to pi H X times two minus three over our x. Solving for X. We have pi H X equals zero or You have to -3 over our x equals zero. You get x equals zero or X. Which is 2/3 are so the critical points are Mexico zero and 2/3 are now that we only have limited values for X and Y. Since they're both measures of length and we know that X and Y are All greater than zero. And since why is equal to h minus H over our X. Then this should be mhm greater than or equal to zero. And solving for X. We have minus H over our x. Greater than or equal to negative H. Or that's ex vessel or equal to are. So this tells us that the domain of V must be Between zero and are only. And because we have a particular integral and a critical point then we will apply the concept of absolute extreme. That means to find the absolute maximum. We need to evaluate the function V at the end points and at the critical point. So we have V. Of zero. That is pi H time zero squared times H minus H. Over our Time zero that will be zero. And if our is two or X. S two thirds are we have V. Of two thirds are that's by times age times 2/3 are squared times H minus H. Over our times two thirds are this will be 4/9 pi H R squared Times 1 3rd Age. which gives us four over 27 pi times H squared R squared. And if exits are we have V. Of our that's going to be pi H times R squared times H minus H. Over our times are that's by H R squared times H minus H. That will be zero. And so the largest possible volume would be for over 27 pi times age square times R squared. Yeah. Yeah

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