00:01
In this problem we need to show that all rectangle inscribed in a giving circle, square has the maximum area.
00:08
So for this purpose, i have just created a rough diagram and rectangle is a, b, c, d and its sides would be 2x and 2y.
00:21
Also, radius of the circle is, radius of the circle is nothing but a.
00:27
Now if we take this triangle that is a -o and this is suppose this is m.
00:39
So in triangle a -m -o we can use pythagodosuram.
00:44
So x -square plus y -square equals to a -square we can write easily.
00:51
Now if we talk about y, so y is nothing but a square minus x -square and its root.
00:59
Now we can find area and that is nothing but x into y and that is basically or area would be 4xy because side is 2x and 2y.
01:12
So area of rectangle is side into breath into length that is 2x into 2y.
01:18
That is 4xy.
01:20
So this would be 4x and y is nothing but root of a square minus x square.
01:28
Now for the further process we need to find a square because it will give some simplification to us because if a is maximum a square will be maximum.
01:42
A is minima so a square will be the same thing.
01:46
So this would be 16 x square times of a square minus x square basically.
01:55
So if we simplify this now we can consider this as some other function that is b.
02:00
Now we can find dv upon d x and that is that is simply we can multiply this also...