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!

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- May 27th 2008, 05:47 AMsasomextremes of function
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! - May 27th 2008, 07:04 AMsasom
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? - May 27th 2008, 08:26 AMSoroban
Hello, sasom!

Quote:

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: .

Quote:

Where is the minimum and where is the maximum?

Then: .

At

At

. . Since

- May 28th 2008, 12:36 AMsasom
Thank you!