00:01
Alright, so what we have is we have the unit circle, which is this right here, of radius one, right, it's the equation x squared plus y squared equals one, unit circle, and there is a rectangle that's going to be inscribed in this thing, and we want to maximize the area.
00:22
So here's our rectangle, let me use different colors, alright, so there's that, and this right here.
00:33
Okay, so the thing you got to realize is that, like this distance here is y units, this is also y units, but to get to that point it's negative y.
00:46
This distance here is x, but this is also x units, but to get to this point that coordinate is going to be negative x.
00:55
So the area is going to be 2x for the base times 2x, 2y for the height.
01:11
Alright, now to figure out, to maximize the area of this rectangle, we need to take the derivative of that function right there, but we have x's and y's all mixed together, kind of a bummer, but that's okay.
01:31
If i take this equation, x squared plus y squared equals one, and solve it for y, one minus x squared, and then take the square root, y is going to equal one minus x squared, it's really plus and minus one minus x squared, but we can get away with just leaving it positive, it'll be fine.
01:57
So now i can take that and put it in for y, so my area is 2x times the square root of one minus x squared.
02:07
Cool, so now let's take the derivative, a prime, so the derivative of the first is two, times the second thing, plus the 2x times the derivative of that square root, which is going to be negative 2x over the square root of one minus x squared, and there is a two out here, two square root of, because you raise it to the negative half, bring it down, okay good.
02:39
So these twos can reduce out, so a prime is equal to two square root of one minus x squared, minus 2x squared over the square root of one minus x squared, and we are going to set this thing equal to zero, zero, zero...