Closed form for the maximum of the two-variable function $(a x + (1 - a) y) (b (1 - x) + (1 - b) (1 - y))$

I’m trying to characterize the maximum of this function within the unit interval ($x,yin [0,1]$):

$$f(x,y)=(a x + (1 – a) y) (b (1 – x) + (1 – b) (1 – y))$$

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:

$nablamathcal L(x,y,lambda)=0$

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 $y=1-x$.