Constrained optimization of $f(x,y)=xy$ 
Find the constrained extrema of $f(x,y)=xy$ at $x+y=1$.

By the Lagrange multiplier theorem:
$$
\begin{cases} \nabla f(x,y)+ \lambda \nabla g(x,y)=0
\\ x+y=1 \end{cases}
\iff
\begin{cases} y+ \lambda = 0
\\ x+\lambda = 0 \\ x+y=1 \end{cases}
$$ 
Which gives me the condition that $y=x \Rightarrow$ plugging this in the last equation gives me that $(\frac{1}{2}, \frac{1}{2})$ is a critical point.
Now, I don't know how to classify this point. It's evident that it's a maximum, but I don't know how to prove it.
 A: One approach is akin to 1st derivative test in 1D. Take small step away, say $x = \frac12 + \epsilon,y = \frac12 -\epsilon$ and note that 
$$
f\left(\frac12+\epsilon, \frac12 - \epsilon\right)
 = \frac14 - \epsilon^2
 < \frac14
 = f\left(\frac12, \frac12\right),
$$
so indeed $(1/2,1/2)$ is a maximum.

If you are not limited in your choice of technique, it's much easier for that problem to substitute $y=1-x$ and then consider the function
$$
g(x) = f(x,1-x) = x(1-x) = x - x^2,
$$
which is a parabola opening down and hence has no minima and a maximum at
$$
x = \frac{-b}{2a} = \frac{-1}{2\cdot(-1)} = \frac12.
$$
A: A different approach without Lagrange's Theorem:
$f(x,y) = xy = 1/4(x+y)^2 -1/4(x-y)^2 = 1/4 - 1/4(x-y)^2.$
Hence, the maximum of f (global) = 1/4
A: With $y=1-x$, the maximum of $x(1-x)$ occurs at $x=\frac12$ (and $y=\frac12$), with value $\frac14$.
A: We can use a change of coordinate such as a rotation in the plane $xy$,
$$x:=\frac{u+v}{\sqrt2},y:=\frac{u-v}{\sqrt2}$$
and the problem becomes the maximization of
$$\frac{u^2-v^2}2$$ under $\sqrt2u=1$. This obviously occurs with $v=0$, giving the value $\frac14$.
A: Analogously to using the Hessian in an unconstrained optimization problem, you can use the signs of certain minors of the bordered Hessian to classify a constrained extremum. Section 4 of this handout on bordered Hessians in particular applies directly to this problem.
With $L(x,y,\lambda) = xy+\lambda(x+y-1)$ the bordered Hessian works out to be the constant matrix $$\begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix}.$$ There are two variables and one constraint, so we only need to examine the determinant of this matrix, which is equal to $2$. This is greater than zero, so you’ve found a local maximum.  
This particular problem, as are many others of the simpler exercises that are typically given when first learning the Lagrange multiplier method, is more easily solved using other methods, as witnessed by the variety of answers that you’re getting. You can take advantage of those other methods to check your work.
