Conjecture: Given constraint $g(x,y)=0$ that implies $y=f(x)$ and a quantity we wish to maximize(minimize), $h(x,y)$, then, if h(x,y)=h(y,x), an extremum obtains when x=y. If $x>>y$ corresponds with arbitrarily large values of h, then x=y corresponds to a minimum. If $x>>y$ corresponds with arbitrarily small values, $x=y$ corresponds to a maximum.
Fermat addressed the following problem with primordial calculus: Given a line segment of length L, where should you cut it such that, were the resulting pieces to make two of the four legs of a rectangle, the area would be at a minimum?
You can do this without calculus:
Let x+y=L where x and y are to be the length and width of the rectangle, respectively. We wish to minimize xy.
The bigger x, the smaller y. Once x is greater than L/2, it starts taking on values formerly taken by y, and y now obtains those values taken by x. In either case, x+y is the same, and xy is the same, so we only need consider values of x from 0 to L/2.
These considerations suggest we take advantage of this symmetry: Let $$x=\frac{L}{2}-q$$ $$y=\frac{L}{2}+q$$
So, for any value of q, $x+y=L$, and $$A=\frac{L^2}{4}-q^2$$.
Clearly, Area is a maximum when $q=0$.
I think there's a theorem at work here.
I understand how this works with Lagrange multipliers, but I believe this can be proven in only geometric terms, preferably with no explicit reference to the slope of tangent lines. I suspect arguments using area are possible as in the above.
Assuming this theorem is true, then in specific problems, extrema can be found without having to use a second derivative test. If symmetry is sufficiently clear, extrema can be found with a minimum of algebra.
I do not know if this answers your question.
Maximization with geometry.
Let $L$ be fixed, and $x+y=L$.
Draw $|AB| =L.$
Draw the Thales circle at the midpoint $M$ of $\overline {AB}$. Pick any point $C$ on the circle.
Draw the height $h_{AB}$ from $C$ onto $\overline {AB},$ intersecting at point $D.$
Triangle $ABC$ is a right triangle, right angle at $C $.
$D$ divides $L$ into two parts : $|AD|=x|$ and $|DB|=y$;
$x+y=L.$
In the right triangle $ABC$ we have
$h^2=xy$ (rule of altitudes).
$h$ is maximal for $h= |AB|/2=L/2,$ radius of the circle.
Then triangle $ABC$ is isosceles and $x=y.$
$h_{max}^2= L^2/4 $
What is $h_{min}$?