 Hi and welcome to the session. Let us discuss the following question which says show that the height of the cylinder of maximum volume that can be inscribed in a cone of height h is 1 by 3 h. Let's see its solution. First of all in this figure we have inscribed this cylinder in this cone. Now we are given the question that the height of the cone is h so that means here O B is h. Let us assume that the radius of the cone say O A is R and the radius of the cylinder say B G is X. Now we will also assume that the height of the cylinder is Y that is O G is Y. So now we have let Y be the height and X be the radius of the cylinder which is inscribed in a cone of height and whose radius is R. Thus here we have O B is equal to h, O G is equal to Y, O A is equal to radius R and G D is equal to the radius of the cylinder that is X. Now we can clearly notice that triangle B G D and B O A are similar. So we have triangle B G D and triangle B O A are similar. So that means the ratio of the corresponding sides of these two triangles will be equal. So from this we have G D upon O A is equal to B G upon B O and this implies that X upon R is equal to B G that is B O minus G O which will be equal to H minus Y upon B O that is H. So we have X upon R is equal to H minus Y upon H. Now from here we will find out the value of X. So from this we get X is equal to R into H minus Y upon H. Now in question we are given that the cylinder is of maximum volume. So let us assume that V be the volume of the cylinder. Now volume of a cylinder is pi R square H where R is the radius and H is the height of the cylinder. So here the volume of this cylinder will be pi into radius square and the radius of the cylinder is X. So you have X square into height that is Y because height of the cylinder is Y. Thus the volume is pi X square Y. Now let us substitute the value of X from here. So V that is volume is equal to pi into R into H minus Y upon H whole square into Y which will be equal to pi into R square into H minus Y whole square into Y upon H is square. That is pi R square upon H square into H square plus Y square minus 2 H Y into Y which will be equal to pi R square upon H square into H square Y plus Y cube minus 2 H Y square. Now for maxima V dash of Y will be equal to 0. So from here let us find out V dash of Y this will be equal to pi R square upon H square into H square plus 3 Y square minus 4 H Y. Now V dash of Y will be equal to 0 so that means this expression will be equal to 0. So from here we have H square plus 3 Y square minus 4 H Y is equal to 0. Now we will solve this equation to get the value of Y. So this equation can be written as H square minus H Y minus 3 H Y plus 3 Y square is equal to 0. This implies H into H minus Y minus 3 Y into H minus Y is equal to 0 that is H minus Y into H minus 3 Y is equal to 0. So from this we have either H minus Y is equal to 0 or H minus 3 Y is equal to 0 that is either Y is equal to H or Y is equal to 1 by 3 H but we know that the cylinder is inscribed in the cone so that means the height of the cylinder and the height of the cone can never be equal. So we have but Y is equal to H is not possible because cylinder is inscribed in the cone. So from this we have Y is equal to 1 by 3 H and Y is the height of the cylinder so that means height of the cylinder is equal to 1 by 3 H. Now we know that for maximum volume V double dash of 1 by 3 H should be less than 0. So first of all let us find out V double dash of Y this will be equal to pi R square upon H square into 6 Y minus 4 H. Now let us find V double dash of 1 by 3 H so this will be equal to pi R square upon H square into 6 into 1 by 3 H minus 4 H which will be equal to pi R square upon H square into 2 H minus 4 H which is equal to pi R square upon H square into minus 2 H which is equal to minus 2 pi R square upon H and this is less than 0. So that means the volume of this cylinder will be maximum when the height of the cylinder will be 1 by 3 H. Thus volume of the cylinder that is V is maximum when Y is equal to 1 by 3 H. With this we finish this session hope you must have understood the question. Goodbye take care and have a nice day.