maximum of function in bounded area

Question

How can i calculate maximum of $ \frac{-1}{(x+y+3)^{2}} $ in [-1 1]x[-1 1] with non numeric method. I know that -0.2 is maximum of this function with numeric method and The Hesian matrix is zero . fxxfyy-fxyfyx=0

Thanks a lot

Answer 1

Perhaps I don't get it right, but here is a try to simplify your question.

$(x+y+3)^2>0$, hence $ \dfrac{-1}{(x+y+3)^2}$ is maximum when $(x+y+3)^2$ is too.

But very clearly on $[-1,1]\times[-1,1]$, $|x+y|<3$.

Thus $(x+y+3)^2$ is maximum on $[-1,1]\times[-1,1]$ when $(x+y+3)$ is maximum on $[0,1]\times[0,1]$

This is clearly for $x=y=1$

You have then $ max \dfrac{-1}{(x+y+3)^2}=-\dfrac{1}{25}=-0.04$