Let there be a function f, continuous, differenciable, and positive monotonic in variables x and y. For example, let f be
$f(x,y) = (a*x^2 + b*y^2)/2$
, where a and b are constants
, and let us try to max this function with a constraint, take $max(x,y) f, s.t. x+y=1$
Using lagrangian, we need
$∂f / ∂x = ∂f / ∂y$
, which implies $ax=by$
, or $a/b=y/x$
this implies the larger a to b ratio, the smaller the x to y ratio, and is very counterintuitive to me. Should the larger a is the more valuable x is and more goes to x?
Here my solution is $x = b/(a+b)$
The more valuable y is the more should be spend on x? This is neither the min solution
How did I use lagrangian wrong? Or are the preconditions not met? Thanks!
For positives $a$ and $b$ by C-S $$\frac{ax^2+by^2}{2}=\frac{\left(\frac{1}{a}+\frac{1}{b}\right)(ax^2+by^2)}{2\left(\frac{1}{a}+\frac{1}{b}\right)}\geq\frac{(x+y)^2}{2\left(\frac{1}{a}+\frac{1}{b}\right)}=\frac{ab}{2(a+b)}.$$ The equality occurs for $$\left(\frac{1}{\sqrt{a}},\frac{1}{\sqrt{b}}\right)||\left(\sqrt ax,\sqrt by\right)$$ or $$ax=by,$$ which says that we got a minimal value.
The maximum value does not exist.
For $a<0$ and $b<0$ by the similar way we can get a maximal value: $\frac{ab}{2(a+b)}.$