Answer to Question #334437 in Calculus for sandeep

Question

f(x,y)= -(x²-1)²-(x²y-x-1)²

Find the critical points of f (x, y) and Using the Second Partials Test (SPT), find the relative extrema’s of f (x ,y)

Answer 1

$$f(x,y)= -(x^2-1)^2-(x^2y-x-1)^2$$

Find the critical point(s)

$$f_x=-2(x^2-1)(2x)-2(x^2y-x-1)(2xy-1)$$

$$f_y=-2x^2(x^2y-x-1)$$

$$\begin{cases} f_x=0 \\ f_y=0 \end{cases}$$

If $x=0,$ then

$$-2(0-1)(0)-2(0-0-1)(0-1)=-2\not=0$$

If $x^2y-x-1=0,$ then

$$-2(x^2-1)(2x)-2(0)(2xy-1)=0$$

Since $x\not=0,$ we take $x_1=-1, x_2=1.$

$x=-1$

$$(-1)^2y-(-1)-1=0=>y=0$$

Point $(-1, 0).$

$x=1$

$$(1)^2y-1-1=0=>y=2$$

Point $(1, 2).$

$$f_{xx}=-12x^2+2-12x^2y^2+12xy+4y$$

$$f_{xy}=f_{yx}=-8x^3y+6x^2+4x$$

$$f_{yy}=-2x^4$$

Point $(-1, 0)$

$$f_{xx}(-1,0)=-10<0$$

$$D=\begin{vmatrix} -10 & 2 \\ 2 & -2 \end{vmatrix}=16>0$$

There is a relative maximum at $(-1, 0).$

Point $(1, 2)$

$$f_{xx}(1,2)=-26<0$$ $$D=\begin{vmatrix} -26 & -6 \\ -6 & -2 \end{vmatrix}=16>0$$

There is a relative maximum at $(1, 2).$