# Find the local maximum and minimum values, saddle points of f(x,y) = x4 + y4 – 4xy + 1

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Solution

f(x, y)  = x4 + y4 – 4xy + 1

Then

fx = 4x3 – 4y

fy = 4y3 – 4x

Also,

fxx = 12x2       and        fyy = 12y2

For critical Point, set

fx = 0      so, 4x3 – 4y = 0

fy = 0     so, 4y3 – 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 fxx(-1, -1) = 12 > 0   So, (-1, -1) is a local minimum.

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