# Does maximizing an increasing function of two variables in more favorable conditions always increase both inputs?

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.

• Thanks for this. Does the method of Lagrange multipliers have anything to say about how you need to increase (x,y) if you increase K? – Martin Leslie Oct 7 '14 at 16:19
• The relation y=2x must hold. At $K=10$ you have $x=10/3,y=20/3$. For $K=12$, $x=10/3$ has to be raised by $1/\color{blue}{3}\cdot 2$. The "2" is here the increasing of K (=12-10). Thus the new (optimal) value for x is $10/3+2/3=12/3=4$. The denominator is 3 (blue), because you have three parts. One y-part and two x-parts. The new (optimal) value for y is $20/3+2/3\cdot 2=8$ – callculus Oct 7 '14 at 17:01