Hi.
appoint local extremes of function f(x,y)=x^3 + y^3 -3xy... i got two points, one is (0,0) and the other is (1,1) - is that correct? I wonder, where is the minimum and where is the maxium?
Thank you!
Hi.
appoint local extremes of function f(x,y)=x^3 + y^3 -3xy... i got two points, one is (0,0) and the other is (1,1) - is that correct? I wonder, where is the minimum and where is the maxium?
Thank you!
am... i used partial derivate so i got:
df/dx = 3x^2 - 3y
df/dy = 3y^2 - 3x
local ekstrems are:
3x^2 - 3y = 0
3y^2 - 3x = 0
.
.
.
x1 = 0, x2 = 1
y1 = 0, y2 = 1
and then, where is maximum and where is "saddle",... I did secodn derivate and I used Hessian matrix, so in point (0,0) this matrix is negative (-9, so this is saddle right? - is this minimum?) and in point (1,1) a got positive matrix (27 - is this maximum or what?). Is all correct or I did some mistakes?
Hello, sasom!
You found the first partial derivatives, equated them to zero, and solved . . .2. Appoint local extremes of function
i got two points: (0,0) and (1,1) ... is that correct? . . . . Yes!
. .
Substitute [1] into [2]: .
Hence: .
There are two critical points: .
You're expected to know the Second Partials Test . . .Where is the minimum and where is the maximum?
Then: .
At
At
. . Since