Simple? Math problem from a physics textbook Given 
$$\cos A ={ {xy-ab} \over {\sqrt{x^2-b^2}\sqrt{y^2-b^2}}},$$
where $x>0,y>0,x+y=a+b=const$. 
Here $x$ and $y$ are variables , $a$ and $b$ are constants.
They say that the biggest possible $\cos A$ is when $x=y$ cause there is symmetry. Substitution $(x,y)$ to $(y,x)$ gives the same function.
I have not studied two argument functions yet, so for me it is not that clear. Can you show me a simple way to understand it.Thanks.
 A: "They say that the biggest possible $\cos A$ is when $x=y$ cause there is symmetry"
They were lying (twice).
Physicists love symmetry. They love it so much that they are willing to assume all functions are "well behaved" in a sense that symmetry arguments can be applied to them. But  In general, when you have a continuous function $f(x,y)$ which is symmetric in the sense that $f(x,y)=f(y,x)$, then it does not necessarily follow that the maximum or minimum of the function with respect to $x+y=c$ is attained when $x=y$. For an example, consider the function $f(x,y)=(x^2+y^2)\sin(1/(x^2+y^2))$ for $(x,y)\neq (0,0)$ and $f(0,0)=0$. This is a symmetric function but the value zero at point $(0,0)$ is neither a maximum nor a minimum with respect to values along the line $x+y=0$, because clearly in every neighborhood of the origin along the line $x+y=0$ we can find both positive and negative values of $f$. So symmetry is in general insufficient for extremal values, hence they were lying once.
So when is it true? Let's see what do we need for the symmetry argument to work. Suppose you know that $f(x,y)=f(y,x)$ for all points $(x,y)$ along the line $x+y=c$. The point $(c/2,c/2)$ is the unique point of intersection between the line $x+y=c$ and the line of symmetry $x=y$. Suppose now that $(c/2,c/2)$ is neither a local maximum nor a local minimum of $f(x,y)$ along the line $x+y=c$. You would like to be able to say that in some interval drawn along the the line $x+y=c$, the values of $f$ on one side of the center $(c/2,c/2)$ will be smaller than $f(c/2,c/2)$ and the values on the other side of the center will be larger, or perhaps the other way round, but anyway, you would like to have a picture that a Physicist has in mind when told that a function has no local minimum or maximum at some point. (This can be guaranteed by assuming that the function has continuous partial derivatives, for example, which is an assumption any physicist would make without blinking) In that case symmetry kicks in, and you will find a point $(x_1,y_1)$ on one side of the center $(c/2,c/2)$ such that $f(x_1,y_1)>f(c/2,c/2)$, say, whereas it's reflection to the other side of the center, namely the point $(y_1,x_1)$, would satisfy $f(y_1,x_1)<f(c/2,c/2)$, (or the other way round, perhaps), and that's a contradiction to $f(x_1,y_1)=f(y_1,x_1)$.
So symmetry will guarantee extremal values for "well behaved" functions, but even for those, it won't always guarantee a maximum -- it could not of course because you could always multipliy your function by $-1$ -- so there is something more to be said in order to prove that $\cos A$ is actually a maximum when $x=y$. Consequently, they were lying twice.
A: HINT
The problem boils down to a product of 2 identical functions in the same variable, either $xy$ or $\sqrt{x^2-b^2} \sqrt{y^2-b^2}$. Think about what values should $x,y$ take for the product $xy$ to be maximum if $x+y$ is fixed. You will easily see it is when $x=y$. Here is how: $x+y= c$ so $y=c-x$ and you want to maximize
$$
f(x,y(x)) = xy = x(c-x).
$$
It's easy to see the maximum will occur at $x = c/2$, implying $y = c-x = c/2$ as well...
