I'm trying to characterize the maximum of this function within the unit interval ($x,y\in [0,1]$):
$$f(x,y)=(a x + (1 - a) y) (b (1 - x) + (1 - b) (1 - y))$$
for $0 < a < 1$ and $0 < b < 1$.
An observation is that each factor is a convex combination of $x$ and $y$.
By plotting the graph of $f$, I see that there is a unique maximum for any value in the range of $a$ and $b$. I have tried using Lagrange multipliers with no luck using the condition $x+y-2=0$. If $g(x,y)=x+y-2$, the Lagrange function is:
$$\mathcal L(x,y) = f(x,y)+\lambda\,g(x,y)$$
I compute the gradient of $\mathcal L$ and solve the system of equations:
$\nabla\mathcal L(x,y,\lambda)=0$
The solutions are: $$x* = \frac{4a - 3}{2 (2 a - 1)}$$ $$y* = \frac{4a - 1}{2 (2 a - 1)}$$ $$\lambda = \frac{a-b}{2 (2 a - 1)}$$
In order for the condition to be active, $\lambda > 0$, iff, $1>a>b>\frac{1}{2}$ or $\frac{1}{2}>b>a>0$ or $1>b>\frac{1}{2}>a>0$ or $1>a>\frac{1}{2}>b>0$
The objective function evaluated on the solution is: $$f(x*, y*)=\frac{1-2b}{4(1-2a)}$$
which is positive only if $a,b>\frac{1}{2}$ or $a,b<\frac{1}{2}$.
I would conclude that the maximum would be
$$x* = \frac{4a - 3}{2 (2 a - 1)}$$ $$y* = \frac{4a - 1}{2 (2 a - 1)}$$
when $1>a>b>\frac{1}{2}$ or $\frac{1}{2}>b>a>0$. However, what is the expression of the maximum for the other cases of $a$ and $b$?
For, instance, I can clearly see in the plot that when $a=1$ and $b=0$, the maximum is $(0, 1)$ and when $a=0$ and $b=1$ the maximum is $(1, 0)$. And when $a=b=\frac{1}{2}$, the maximum is the line $y=1-x$.