Question

Example 3 find the local maximum and minimum values and saddle points of f(x, y) = x4 + y4 − 4xy + 1. solution we first locate the critical points:

Answer 1

$f(x,y)=x^4+y^4-4xy+1$

$\begin{cases}f_x=4x^3-4y\\f_y=4y^3-4x\end{cases}$

We have critical points whenever both partial derivatives vanish:

$4x^3-4y=0\implies x^3=y\implies x=y^{1/3}$
$4y^3-4x=0\implies y^3=x=y^{1/3}\implies y^6=y\implies y=0,y=-1,y=1$
$\begin{cases}y=0\implies x=0\\y=-1\implies x=-1\\y=1\implies x=1\end{cases}$

The Hessian is

$\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}12x^2&-4\\-4&12y^2\end{bmatrix}$
$\det\mathbf H(x,y)=(12xy)^2-16$

At (0,0), we have $\det\mathbf H(0,0)=-16<0$, so (0,0) is a saddle point.

At (-1, -1), we have $\det\mathbf H(-1,-1)=128>0$ and $f_{xx}(-1,-1)=12>0$, so (-1, -1) is a local minimum.

At (1, 1), we have $\det\mathbf H(1,1)=128>0$ and $f_{xx}(1,1)=12>0$, so (1, 1) is also a local minimum.

Answer 2

At (1,1) and (-1,-1) there is minimum value and at (0,0) is the saddle point.

Given-

Given function is-

$f(x,y)=x^4+y^4-4xy+1$

Locate the critical points,(use the derivation with respect to x and y)

$f'(x)=4x^3-4y\\\\f'(y)=4y^3-4x$

Equate the above partial derivatives to the zero, we get.]

$4x^3-4y=0\\\\x^3=y$

The value of x is obtained. Solve the equation two and put this value of x in that equation, we get,

$4y^3-4x=0\\y^3-x=0$

Put the value of x, we get,

$x^9-x=0$

$x(x^8-1)=0$

from above equation we get one value of $x$ is zero.

$x^8-1=0$

To the above equation we can rewrite,

$x^2-1=0$

$(x-1)(x+1)=0$

hence we get three values of the $x$ (0,1,-1)

From the equation of partial derivative equation, we get that

$x=0; y=0$

$x=1;y=1$

$x=-1;y=-1$

Thus three critical points are (0,0),(1,1) nd(-1,-1). Now calculate second partial derivative and D(x,y).

$f"(x)=12x^2-0$

$f'(xy)=-4$

$f"(y)=12y^2-1$

$D(x,y)=f"(x)f"(y)-f'(xy)^2$

$D(x,y)=144x^2y^2-16$

Check at critical points,

$D(0,0)=-16<0$

thus no local minima or maxima at point (0,0). this is the saddle point.

$D(1,1)=12>0$

so (1, 1) is a local minimum.

$D(-1,-1)=12>0$

so (-1, -1) is a local minimum.

Hence at (1,1) and (-1,-1) there is minimum value and at (0,0) is the saddle point.

For more about the maxima and minima follow the link given below-

https://brainly.com/question/6422517