Respuesta :
Answer:
The points are "(2,2)".
Step-by-step explanation:
Given value:
[tex]f(x,y)=x^3-6xy+y^3+7[/tex]
Finding the first partial derivation which is equal to 0.
[tex]\to f_x(x,y)=3x^2-6y...............(a)\\\\ \to f_y(x,y)=-6x+3y^2...................(b)[/tex]
solve both the above equation by equal to 0.
For equation (a)
[tex]\to 3x^2-6y=0\\\\\to 3x^2=6y\\\\\to x^2=2y\\\\\to y=\frac{x^2}{2}\\\\[/tex]
For equation (b)
[tex]\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\\\\[/tex]
by solving the above value in the form of x we get:
[tex]\to x=0\\ \to x=2\\ \to x= -1 +\sqrt{3} i\\ \to x= -1 -\sqrt{3} i\\[/tex]
by solving the above value in the form of y we get:
[tex]\to y=0\\ \to y=2\\ \to y= -1 +\sqrt{3} i\\ \to y= -1 -\sqrt{3} i\\[/tex]
Solve the value by applying the second derivation method:
[tex]\to f_{xx}(x,y)=6x...............(a1)\\\\ \to f_{yy}(x,y)=6y...................(b2)\\\\\to f_{xy}(x,y)=-6...................(c2)[/tex]
calculating the value of discriminate:
[tex]d=f_{xx}(x,y) f_{yy}(x,y)-[f_{xy}(x,y)]^2[/tex]
The critical point of the given equation will be (2,2)
[tex]d=f_{xx}(2,2) f_{yy}(2,2)-[f_{xy}(2,2)]^2\\\\[/tex]
[tex]= (6 \times 2) (6 \times 2) -[(-6)^2]\\\\= (12) (12) -[36]\\\\= 144 -36\\\\= 108\\\\[/tex]
Answer:
[tex](2,2)[/tex] is a minimum point of the function [tex]f(x,y).[/tex]
Step-by-step explanation:
[tex]f(x,y)=x^3-6xy+y^3+7[/tex]
Find the partial differentiation w.r.t. [tex]x[/tex] and [tex]y[/tex].
[tex]\frac {\partial f}{\partial x}=3x^2-6y[/tex]
[tex]\frac {\partial f}{\partial y}=-6x+3y^2[/tex]
[tex]\frac {\partial^2 f}{\partial x^2}=6x[/tex]
[tex]\frac {\partial^2 f}{\partial y^2}=6y[/tex]
[tex]\frac{\partial f}{\partial xy}=-6[/tex]
Find the critical points.
[tex]\frac {\partial f}{\partial x}=\frac {\partial f}{\partial y}=0[/tex]
[tex]\Rightarrow 3x^2-6y=0 \;\text{and}\; -6x+3y^2=0[/tex]
[tex]\Rightarrow 6y=3x^2\Rightarrow y=\frac{x^2}2[/tex]
[tex]\Rightarrow -6x+\frac 34 x^4=0\Rightarrow \frac {x^3}4=2[/tex]
[tex]\Rightarrow x^3=8\Rightarrow x=2,y=2[/tex]
Therefore, [tex](2,2)[/tex] is a critical point.
Now, [tex]\frac {\partial^2 f}{\partial x^2}\frac {\partial^2 f}{\partial y^2}-{\frac {\partial f}{\partial xy}}^2[/tex]
[tex]=6x6y-36=36xy-36>0 \;\text{at critical point }\; (2,2)[/tex]
Thus, [tex](2,2)[/tex] is not a saddle point.
[tex]\because \frac{\partial^2 f}{\partial x^2}>0 \; \text{and}\; \frac {\partial^2 f}{\partial x^2}\frac {\partial^2 f}{\partial y^2}-{\frac {\partial f}{\partial xy}}^2>0 \;\text{at point}\; (2,2)[/tex]
Hence, [tex](2,2)[/tex] is a local minimum of a given function.
