00:01
For this problem, we have this illustration of the isosceles triangle of least area that circumscribes a circle of radius r.
00:13
Now the objective here is that have a least area of the isosceles triangle.
00:20
So least a which equals one half base times height or that's the same as b over two times h.
00:29
Now in this isosceles triangle, let's call this our height h and then this should be our base b.
00:38
So this part will be half of the base and then since this one starts from the center of the circle to a point on the circle, then this is also r.
00:51
This one is h minus r and this one using pythagorean should be the square root of h minus r squared minus the square of r or r squared or that's the same as the square root of h squared minus two times r times h.
01:15
Now by similar triangles, base over the height that is b over two over h should equal r over the height which is the square root of h squared minus two r h.
01:36
In other words, b over two equals r times h over square root of h squared minus two r times h.
01:46
Thus we can rewrite our area function in terms of h.
01:51
That'll be a equal to r h over square root of h squared minus two times r times h times h or that's the same as r h squared over square root of h squared minus two times r times h.
02:10
And then we want to get the derivative of a with respect to h...