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Find the local extrema of the function f(x,y) = x^4 + y^4 - 2(x - y)^2 where f(x,y)≤ 0.

My attempt:

from the partial derivatives,

f_x = 4x^3 - 4(x-y) = 0

f_y = 4y^3 + 4(x-y) = 0

we get (x,y) = (0,0),(√2,-√2),(-√2,√2) (lets name these critical points O,A,B)

Second derivatives,

f_xx= -4, f_xy= 4, f_yy= -4

Expanding the function using taylor series at the origin O = (0,0),

1/2.[f

_{xx}(0,0)x

^{2}+

*2*.f

_{xy}(0,0)xy + fyy(0,0)y

^{2}] = -2(x-y)^2

From the definition:

For all values of (x,y) near (0,0),

f(x,y)≥f(0,0) → At (0,0) f(x,y) has a local minimum

f(x,y)≤f(0,0) → At (0,0) f(x,y) has a local maximum

What I understand is that from the taylor expansion, we know that f(0,0) is always negative (or 0).In other words, f(x,y) is always negative(or 0) at (0,0). Right?And since at the beginnging of the problem, the constraint f(x,y) ≤ 0 is given, it is also negative or 0 but how to know that f(x,y) bigger or than f(0,0)? So lets say there are points

P=(-1,-2) and Q = (1,2). Subtituting these I get -2 and no matter what values I put,they're always gonna be negative and that means they're always gonna be maximum because it can never go beyond 0,right? I am going to apply what I think and please correct me if I am wrong.

The function at O is -2(x-y)^2.

Substituting values of x and y with 0,

so f(0,0) = 0 and

substituting values of x and y with 1 and 2,

so f(1,2) = -2 and

thus, f(1,2) is lower than f(0,0). Generalizing it, we have f(x,y) lower than or equals to(when we substitute with 0) f(0,0) and so, f(x,y) has a local maximum value of 0 at (0,0).