Extremum of $f (x,y) := x^2+y^2$ subject to the equality constraint $x+y=3$

Question

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?

Answer 1

By CS inequality: $$ x+y=(x,y)\cdot (1,1)\le \sqrt{x^2+y^2}\sqrt{2} $$ Since $x+y=3$: $$ 3\le \sqrt{2}\sqrt{x^2+y^2} $$ Now, squaring both sided yields $$ 9\le 2(x^2+y^2) $$ In other words $$ x^2+y^2\ge \frac{9}{2} $$ This lower bound is attained when $x=y=3/2$.

Answer 2

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.