Lagrange — maximum of xy subject to x+y=4

Setting $\nabla f = \lambda \nabla g$ gives $(y, x) = \lambda(1, 1) \rightarrow y = \lambda$ and $x = \lambda$, so $x = y$. The gradient condition equates the two partial derivatives.

Substituting $x = y$ into $x + y = 4$ gives $2x = 4 \rightarrow x = \mathbf{2}$, $y = \mathbf{2}$. The critical point is $(2,2)$.

$f(2,2) = 2 \times 2 = \mathbf{4}$. This is the maximum value of $xy$ subject to the constraint $x+y=4$.