00:01
In this question we are given a situation where we are rotating a right angle triangle triangle so first we will see a right angle triangle like this if this is our right angle triangle triangle so here the hypothesis is given as 12 cm and it is rotating around one of its leg and we will get a design of cone then so here what will be the possible largest volume that we have to state here.
00:41
So we will take our perpendicular as h and the radius as r.
00:46
So when we rotate this triangle with respect to the radius like this, then we will get a cone here and this will be the figure of the code.
01:08
So h is here, r is here.
01:10
So we have to find the volume of comb and now hypoteness is at the same place so let's solve this question so as for the question question we will take here pythagoras theorem statement which is perpendicular square plus base square is equal to hypotenous square so a value for hypothesis 12 square base is h square perpendicular is h square and base is r square so a value in the terms of r square will be 144 minus h square this will be our value for base and now volume of cone of formula is 1 by 3 pi r square h so let's put down here our value so 1 by 3 pi r square h so r square is 144 minus h squared multiplied by h.
02:22
So when we simplify this, this will be 1 by 3 by 144 h multiplied minus h q.
02:32
And now this is our volume and after that we will differentiate the previous equation with respect to h.
02:44
So we will get dv by d h as equal to pi by 3, 144 minus 3 h squared.
02:57
When we again differentiate, we will get d square b, d h squared is equal to pi by 3 within the bracket minus 6 h...