I need calculate the absolute end point of the function $f(x,y)=x^2y^3(1-x-y)$ when $K=\{(x,y)\epsilon\mathbb{R}^2: |x|+|y| \leq 1 \}$. So $K$ is a rotated square, how I have to calculate that I started calculating the relative point, but when I try calculate the absolute point I don't have idea how I can do it.
To the absolute point: I consider $F: \mathbb{R}^3 \rightarrow\mathbb{R} / \ F(x,y,\lambda)=f(x,y)+\lambda(|x|+|y|-1)$ but the lagrange system is impossible to solve, I need help! thanks
Firstly, find the extremae of $f$ in the interior of $K$. Then look at the boundary of $K$. In neither of the four cases multipliers are needed as you can solve either equation for $y$ and substitute the solution back in $f$. Don't forget to check the edges.
Lagrange is needed if it's difficult or impossible to solve the restriction for one variable.