Find infimum and supremum of the function $f(x,y)=x^3+y^3$ in $\{(x,y) \in \mathbb R^2 : x^2+y^2 \le 1 \}$.
From the Lagrange theorem: if the infimum and supremum exist then: \begin{align} 3x^2&=\alpha x \tag{1}\\ 3y^2&=\alpha y\tag{2} \\ x^2+y^2 &\le 1 \tag{3}\end{align}
Then: \begin{align} 3x^3&=\alpha x^2 \tag{1}\\ 3y^3&=\alpha y^2\tag{2} \\ x^2+y^2 &\le 1 \tag{3}\end{align}
S0: $$x^3+y^3 =\frac{\alpha}{3}(x^2+y^2) \le \frac{\alpha}{3}$$
But I don't know what I can do to find inf and sup.
If $x^2 + y^2 \leq 1$ then $|x|, |y| \leq 1$, and therefore $$|x^3 + y^3| \leq |x^3| + |y^3| = |x|^3 + |y|^3 \leq x^2 + y^2 \leq 1.$$
It follows that $-1 \leq x^3 + y^3 \leq 1$, for which the bounds are attained by $(\pm 1, 0)$.