00:01
Find the maximum and minimum values of this function.
00:02
So first of all, we've got to take the derivatives, and we're going to set those equal to zero.
00:28
The second equation tells us that y is minus x.
00:35
I'm going to substitute that into the first equation.
00:38
So we've got so that x is zero or two -thirds.
01:11
And then because y is minus x gives us two critical points one is 0 0 the other is 2 thirds minus 2 thirds all right so let's take the second derivatives also equals 8 so our discriminant that's 8 times 6x minus 6 minus the square root of 8 is 64.
02:36
Alright.
02:37
So now we've got two critical points.
02:43
Start with 0, 0.
02:47
So then we got d is 48 minus 64.
02:56
That comes out to be negative 16.
03:03
That's a saddle point.
03:08
And then the other one is two -thirds minus two -thirds.
03:23
Let's see.
03:25
So that's our minimum.
03:49
That is a local minimum, okay? large negative values of x will generally give us values that are smaller than that, more negative, that is...