Solving the following system of equation

Question

I was trying to solve the following problem

Find the extreme values of the function $$f(x,y)=2(x-y)^2-x^4-y^4$$

My attempt- $$p=\frac{\delta f}{\delta x}=4(x-x^3-y)$$ and $$q=\frac{\delta f}{\delta y}=4(y-y^3-x)$$ Equating p and q to 0 I get $$x-x^3-y=0$$ and $$y-y^3-x=0$$ I have no idea on how to solve the above 2 equations

Answer 1

We next simultaneously solve these equations to determine the potential critical points.

We have:

$$x-x^3-y=0 \\ y-y^3-x=0$$

We can write the second equation as $x = y-y^3$ and substitute into the first and get:

$$ y^3(y^2-2)(y^4-y^2+1) = 0 \rightarrow y = 0, \pm~ \sqrt{2}$$

We do not care about the four complex roots.

Next, substitute back into the original equation to find $x$ and this leads to the following potential critical points for test:

$$(x, y) = (0, 0), (-\sqrt{2},\sqrt{2}), (\sqrt{2},-\sqrt{2})$$

Can you proceed?

Answer 2

You can find a good introduction on how to find max or min (or saddles) for a function in two variables here.

Read that. If you still have questions don't hesitate to ask again here.

EDITED: the OP asked how to wsolve the equations. So...

You get two equations. You get $y$ from the first (for example) and you substitute in the second. You will get following equations: $$ x-x^3-(x-x^3)^3=x $$ this can be simplified and factorized in this way: $$ x^3(x^2-2)(x^4-x^2+1)=0 $$ Now you can easily find the solutions can you? Those are just quadratic equations. For you to check the solutions will be $x=0, \sqrt 2, -\sqrt 2$. Then you can take your $x$ and substitute them in your questions to get $y$. This should get you going and you should be able to finish your exercise.

Answer 3

Solve for $y$ from $p=0$ and substitute in $q=0$. You will get $$ x^3\,\left(x^2-2\right)\,\left(x^4-x^2+1\right) =0 $$ For each solution of $x$, find $y$ from $p=0$.

additional work Here are the answers $$x=-\sqrt{2} , y=\sqrt{2} \\ x=\sqrt{2} , y=-\sqrt{2} \\ x=0 , y=0 $$