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Solution

f(x, y) = x^{4} + y^{4} – 4xy + 1

Then

f_{x} = 4x^{3} – 4y

f_{y} = 4y^{3} – 4x

Also,

f_{xx} = 12x^{2 }and f_{yy} = 12y^{2}

For critical Point, set

f_{x} = 0 so, 4x^{3} – 4y = 0

f_{y} = 0 so, 4y^{3} – 4x = 0

Solving these equations we get y = 0, -1, 1 and y = 0, -1, 1

The Hessian is

D = \(\begin{vmatrix}

f_{xx} & f_{xy}\\

f_{yx} & f_{yy}

\end{vmatrix}\)

\(= \begin{vmatrix}

12x^2 & -4\\

-4 & 12y^2

\end{vmatrix}\)

D = (12xy)^{2} – 14

At (0, 0) we have D = -14 < 0 So, (0, 0) is a saddle point.

At (-1, -1), We have D = 128 > 0 and f_{xx}(-1, -1) = 12 > 0 So, (-1, -1) is a local minimum.

At (1, 1) We have D = 128 > 0 and f_{xx}(1, 1) = 12 > 0 So, (1, 1) is a local minimum.

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## Discussion