Extremum of $f (x,y) := x^2+y^2$ subject to the equality constraint $x+y=3$ I had to find the extremum of $z=x^2+y^2$ subject to the constraint $x+y=3$. I used Lagrange multipliers to reach the conclusion that $(1.5,1.5)$ is an extremum point, but had no way of determining whether it's a maximum or a minimum (we did not study the Sylvester criteria). Regardless, intuitively, the most symmetric sum usually gives the largest result, and this is what I used as a justification for the point being a maximum. This is, of course, hardly a mathematical way of showing the correctness of a statement, which is why I ask here what way there is to show it's a maximum in a correct well defined orderly fashion?

Replace $x=3-y$ to the equation
$$z=x^2+(3-x)^2=2x^2-6x+9$$
Take the first derivative to find the critical point(s)
$$\frac{dz}{dx}=4x-6=0 \rightarrow x_c=\frac32$$
Find the second derivative at $x_c$
$$\frac{d^2z}{dx^2}=4$$
Because second derivative is positive the critical point is a minimum.
