00:02
We're asked to find dimensions of the rectangle of largest area that can be inscribed in a circle of radius r.
00:09
So we have a circle of radius r, and we have a rectangle inside of this of maximum area.
00:34
Now, we have the area, such a rectangle.
00:47
If we call the length of one side x and the other side y is going to be x times y.
00:58
Now notice that we can also draw a point from the center of the circle to one of the corners of the rectangle, which has a length between them, a segment of length r.
01:10
So by pythagorean theorem, we have that half of the length of one side, x over 2 squared, plus half of the length of the other side, y over 2 squared, is equal to hypotenuse r squared.
01:28
In this way we have a relationship between x and y.
01:35
Now to help us, notice that a function area is always greater than equal to zero.
01:54
Well, we can't quite do that.
01:58
But notice that both x and y have to be greater than zero.
02:03
So we can solve for y in this equation.
02:06
We have y over 2 squared equals r squared minus x over 2 squared, so that y.
02:14
Over is equal to 2 times the positive square root of r squared minus x over 2 squared and therefore the area as a function of x is x times 2 times the square root of r squared minus x over 2 squared so now we want to find the maximum value of a notice that x and y are both less than r.
03:01
So to do this, i'm going to find the derivative.
03:04
A prime of x, this is going to be, well, before we even do that actually, notice that a is of course always greater than or equal to zero on zero to r and that a therefore has a maximum when the function f, which is a squared, is maximized.
03:44
Now f of x, this is equal to 4 x squared times r squared minus x over 2 squared.
04:00
Now, define the maximum of f.
04:04
F prime of x is, by the product rule, this is 2 times 4, 8x times r squared minus x over 2 squared, plus 4x squared times the derivative.
04:18
The second factor.
04:20
This is negative 2 x over 2 times 1 half...