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 $\displaystyle f(x,y)\:=\:x^3 + y^3 -3xy$
i got two points: (0,0) and (1,1) ... is that correct? . . . . Yes!
. . $\displaystyle \begin{array}{cccccccc}
f_x &=& 3x^2 - 3y &=& 0 & \Rightarrow & y \:=\:x^2 & {\color{blue}[1]}\\
f_y &=& 3y^2 - 3x &=& 0 & \Rightarrow & x \:=\:y^2 & {\color{blue}[2]}
\end{array}$
Substitute [1] into [2]: .$\displaystyle x \:=\:(x^2)^2 \quad\Rightarrow\quad x^4 - x \:=\:0 \quad\Rightarrow\quad x(x^3-1) \:=\:0$
Hence: .$\displaystyle x \:=\:0,1\quad\Rightarrow\quad y \:=\:0,1$
There are two critical points: .$\displaystyle (0,0,0) \text{ amd }(1,1,-1)$
You're expected to know the Second Partials Test . . .Where is the minimum and where is the maximum?
$\displaystyle f_{xx} \:=\;6x\qquad f_{yy} \:=\:6y\qquad f_{xy} \:=\:-3$
Then: .$\displaystyle D \;=\;\left(f_{xx}\right)\left(f_{yy}\right) - \left(f_{xy}\right)^2 \;=\;(6x)(6y) - (-3)^2\;=\;9(4xy-1)$
At $\displaystyle (0,0)\!:\;\;D \:=\:9(0-1) \:=\:-9\quad\hdots\;\;\boxed{\text{saddle point at }(0,0,0)}$
At $\displaystyle (1,1)\!:\;\;D \:=\:9(3) \:=\: +27\quad\hdots\text{ extreme point}$
. . Since $\displaystyle f_{xx}(1,1) \:=\:+6\quad\hdots\boxed{ \text{minimum at }(1,1,-1)}$