Consider the problem of maximizing $\sqrt{x}y$ such that $x+y=10$. By basic calculus we can show that the maximum occurs at $x=10/3$, $y=20/3$. If we loosen the constraint to $x+y=12$ then the maximum now occurs at $x=4$, $y=8$. Notice that we have increased both input values. This makes sense to me: there's some kind of rectangle in my mind and we don't want to make it less square as we increase the size.
What's a more general theorem along these lines? I feel like it should be known in mathematical economics perhaps.
Edit: The kind of theorem I'd like might go something like as follows. Assume $F(x,y)$ is increasing (i.e. $x_1>x_2$ and $y_1>y_2$ implies $F(x_1,y_1)>F(x_2,y_2)$). If $(x_1^*,y_1^*)$ are the inputs that maximize $F(x,y)$ such that $G(x,y)=k_1$ for some increasing function $G$ and $(x_2^*,y_2^*)$ are the inputs that maximize $F(x,y)$ such that $G(x,y)=k_2$ for some $k_2>k_1$ then $x_2^*>x_1^*$ and $y_2^*>y_1^*$.
I'm struggling with formulating a conjecture and with what methods of proof are available.
The method of Lagrange multipliers can help. Here it is:
$\mathcal L=x^{1/2}\cdot y+\lambda \left( 10-x-y \right)$
The first order conditions
$\frac{\partial \mathcal L}{\partial x}=\frac{1}{2}\cdot x^{-1/2}\cdot y-\lambda=0 $
$\Rightarrow \frac{1}{2}\cdot x^{-1/2}\cdot y=\lambda \quad (1)$
$\frac{\partial \mathcal L}{\partial y}=x^{1/2}-\lambda=0$
$\Rightarrow x^{1/2}=\lambda \quad (2)$
$\frac{\partial \mathcal L}{\partial \lambda}=10-x-y=0 \quad (3)$
(1) divided by (2) gives $\frac{1}{2}\frac{y}{x}=1$
Thus $y=2x$.
This is the result you already have. To calculate the the values for x and y you use equation (3).
For the (max) function you get $f\left(\frac{10}{3},\frac{20}{3}|K=10 \right)\approx 12.172$ and $f(4,8|K=12)=16$
From equation (1) or (2) you can calculate $\lambda$ for $(x,y)=\left( \frac{10}{3},\frac{20}{3}\right)$ It is $\lambda \approx 1.826$
The interpretation of the value of Lambda is like the following:
If you increase K (here 10) about $\Delta \ K$, the value of the function increases about $1 .826 \cdot \Delta \ K$
In your case $\Delta K =2$. And $12.172+2\cdot 1 .826=15.824$
It is a little bit under 16. It is only an approximation. The smaller $\Delta K$ is, the more exact is the calclulation with Lambda.
This is what I can say.