How to show that the maximum depends on a given parameter How to show that the following maximization problem has a unique solution, that depends on parameter $a$ (without using differential calculus): $f(x, y)=y-\frac{1}{x}$ subject to $a x+y \leq 1, x \geq 0, y \geq 0$, where $a>0$?
We know that it must be shown that the constraint is a continuous, compact-valued correspondence. Because $f$ is continuous and strictly quasiconcave and the maximum problem has a unique solution. The Theorem of the Maximum implies that this solution depends continuously on $a$. However, we would like to prove it mathematically. Can you help us?
 A: $$\begin{cases} f(x,y)=y-\frac 1x\\ ax+y\leq1\end{cases}$$
We may write:
$$ax+y-\frac 1x \leq 1-\frac 1x$$
Or:
$$f(x,y)=y-\frac 1x \leq 1-\frac 1x-ax=\frac{x-1-ax^2}x$$
So  $f(x, y)$ is maximum when RHS is maximum.Let:
$u=\frac{x-1-ax^2}x$
$u'=\frac{x(1-2ax)-(x-1-ax^2}{x^2}$
$u'=\frac{1-ax^2}{x^2}=0$
which gives:
$x=\pm \frac 1{\sqrt a}$
1): $x=+\frac 1{\sqrt a}$
From inequality we have:
$a\cdot(\frac 1{\sqrt a})+y\leq 1$
For maximum we consider equality $ax+y=1$ and check:
$y=-a\cdot (\frac 1 {\sqrt a})+1=1-\sqrt a$
putting in $f(x, y)$ we get:
$f(\frac 1{\sqrt a}, 1-\sqrt a)= \sqrt a -1-\sqrt a=-1$
This is minimum of $f(x, y)$
2): $x=-\frac 1{\sqrt a}$, we have:
$a\cdot(-\frac 1{\sqrt a})+y-1=0$
Which gives:
$y=1+\sqrt a$
Putting in $f(x, y)$ we obtain:
$f(-\frac 1{\sqrt a}, 1+\sqrt a)=1+\sqrt a+\sqrt a=1+2\sqrt a$
This  is maximum of $f(x, y$ which is a function of parameter $a$.
Using calculus we get:
$\begin {cases}\frac 1{x^2}+ a \lambda=0\\1+\lambda=0\\ax+y-1=0\end {cases}$
Which gives:
$\lambda=-1$, $x=\pm \frac 1{\sqrt a}$ and $y=\sqrt a-1$ and $y=1+\sqrt a$ which are similar results.
