Calculus III local extrema?

Question

I am stuck on this problem: 
I need to find any local extrema/saddle points 
I was given f(x,y)=5-x^4y^4 

Answer 1

The function f(x, y) = 5 - x⁴y⁴.

Because fx (x, y) = - 4x³y⁴ and fy (x, y) = - 4x⁴y³ both partial derivatives are 0 if x = 0 or y = 0. That is, every point along the x- or y-axis is a critical point.

Moreover, because fxx (x, y) = - 12x²y⁴, fyy (x, y) = - 12x⁴y² and fxy (x, y) = - 16x³y³.

If either x = 0 or y = 0, then d = fxx (x, y) fyy (x, y)  - [ fxy (x, y) ]² = (- 12x²y⁴)(- 12x⁴y²) - [ - 16x³y³ ]² = 0. So, the second partials test fails.

The second partials test can fail to find relative extrema in two ways. If either of the first partial does not exist, then cannot use the test. also, if d = 0 the test fails. In such cases as follows given below:

However, because f(x, y) = 0 for every point along the x- or y-axis and

for all other points x⁴y⁴ > 0

                           - (x⁴y⁴) < 0

                          5 - (x⁴y⁴) < 0 + 5

                           f(x, y) < 5.

The function f(x, y) < 5 for all other points, you can conclude that each of these critical points yields an absolute maximum.

Answer 2

The function f(x, y) = 5 - x⁴y⁴.

Because fx (x, y) = - 4x³y⁴ and fy (x, y) = - 4x⁴y³ both partial derivatives are 0 if x = 0 or y = 0. That is, every point along the x- or y-axis is a critical point.

Moreover, because fxx (x, y) = - 12x²y⁴, fyy (x, y) = - 12x⁴y² and fxy (x, y) = - 16x³y³.

If either x = 0 or y = 0, then d = fxx (x, y) fyy (x, y)  - [ fxy (x, y) ]² = (- 12x²y⁴)(- 12x⁴y²) - [ - 16x³y³ ]² = 0. So, the second partials test fails.

The second partials test can fail to find relative extrema in two ways. If either of the first partial does not exist, then cannot use the test. also, if d = 0 the test fails. In such cases as follows given below:

However, because f(x, y) = 0 for every point along the x- or y-axis and

for all other points x⁴y⁴ > 0

                           - (x⁴y⁴) < 0

                          5 - (x⁴y⁴) < 0 + 5

                           f(x, y) < 5.

The function f(x, y) < 5 for all other points, you can conclude that each of these critical points yields an absolute maximum.

Let f(x, y) have continuous second partial derivatives on an open region containing a point 9a, b) for which fx (a, b) = 0 and fy (a, b) = 0.

To test for relative extrema of f(x, y), consider the quantity d = fxx (x, y) fyy (x, y)  - [ fxy (x, y) ]².

1. If d > 0 and fxx (x, y) > 0, then f(x, y) has a relative minimum at (a, b).

2. If d > 0 and fxx (x, y) < 0, then f(x, y) has a relative maximum at (a, b).

3. If d < 0, then (a, b, f(a, b)) is a saddle point.

4. The test is inconclusive if d = 0.