Mathematics: Examine the function f(x,y) = x^3 - 6xy + y^3 + 7 for relative extrema and saddle points.

Examine the function f(x,y) = x^3 - 6xy + y^3 +

kenzi352

02.02.2022

The points are "(2,2)".

Step-by-step explanation:

Given value:

f(x,y)=x^3-6xy+y^3+7

Finding the first partial derivation which is equal to 0.

$\to f_x(x,y)=3x^2-6y...............(a)\\\\ \to f_y(x,y)=-6x+3y^2...................(b)$

solve both the above equation by equal to 0.

For equation (a)

$\to 3x^2-6y=0\\\\\to 3x^2=6y\\\\\to x^2=2y\\\\\to y=\frac{x^2}{2}\\\\$

For equation (b)

$\to -6x+3y^2 =0 \\\\\to 3y^2 =6x\\\\\text{put the value of y in equation b}\\\\\to 3(\frac{x^2}{2})^2 -6x=0\\\\\to \frac{3}{4} x^4 -6x=0\\\\$

by solving the above value in the form of x we get:

$\to x=0\\ \to x=2\\ \to x= -1 +\sqrt{3} i\\ \to x= -1 -\sqrt{3} i\\$

by solving the above value in the form of y we get:

$\to y=0\\ \to y=2\\ \to y= -1 +\sqrt{3} i\\ \to y= -1 -\sqrt{3} i\\$

Solve the value by applying the second derivation method:

$\to f_{xx}(x,y)=6x...............(a1)\\\\ \to f_{yy}(x,y)=6y...................(b2)\\\\\to f_{xy}(x,y)=-6...................(c2)$

calculating the value of discriminate:

$d=f_{xx}(x,y) f_{yy}(x,y)-[f_{xy}(x,y)]^2$

The critical point of the given equation will be (2,2)

$d=f_{xx}(2,2) f_{yy}(2,2)-[f_{xy}(2,2)]^2\\\\$

$= (6 \times 2) (6 \times 2) -[(-6)^2]\\\\= (12) (12) -[36]\\\\= 144 -36\\\\= 108\\\\$

0,0(0 оценок)