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

Answers

This answer is not selected as best answer. This answer may not be sufficient for exam.

Your limit has been exceed. We have implemented this system because, We got difficulty on managing our servers. Please donate some amount to remove this limit.

Quota: 0 / 30

Donate

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.

If you found any type of error on the answer then please mention on the comment or submit your new answer.
Leave your Answer:

Click here to submit your answer.

Discussion
0 Comments
  Loading . . .