I have an optimization problem in the form:
$$\max (a-\bar{a})(b-\bar{b}) \qquad \text{subject to} \qquad a+b=1.$$
Here $\bar{a}$ and $\bar{b}$ are known values and both of them are positive.
Let $a_{opt}$ and $b_{opt}$ are the solutions or optimal values. Then
$$a_{opt}-\bar{a}>0 \qquad \text{and} \qquad b_{opt}-\bar{b}>0$$
How to find this solution for this optimization problem?
Consider that you look for the extremum of function $$F=(a-A)(b-B)$$ subjact to $a+b=1$. So, using $b=1-a$, $$F=(a-A) (1-a-B)$$ $$F'=-2 a+A-B+1$$ So, the first derivative cancels for $a=\frac{A-B+1}{2} $ for which $F=\frac{(A+B-1)^2}{4} $.
But, the second derivative is $F''=-2$ which confirms that this a maximum.
Now, the problem is : $a >0$ ?
Solve it using Lagrange's multipliers and you'll get the answer.
The level lines $\gamma_c:=f^{-1}(c)$ of the function $$f(x,y):=(x-\bar a)(y-\bar b)$$ are hyperbolas centered at $(\bar a,\bar b)$, and having asymptotes $x=\bar a$, resp., $y=\bar b$ (see the following figure). We are looking for the largest $c$ such that $\gamma_c$ meets the red line $\ell: \>x+y=1$. This is the case when $\gamma_c$ just touches the line $\ell$, because a $\gamma_c$ intersecting $\ell$ in two points leaves room for improvement. Due to symmetry this will take place at the point $(x_*,y_*)$ where the $45^\circ$ line $$\sigma:\quad t\mapsto (x,y):=(\bar a+t,\bar b+t)$$ through $(\bar a,\bar b)$ intersects $\ell$. One obtains the value $t_*={1\over2}(1-\bar a-\bar b)$. This leads to $$x_*=\bar a+t_*={1\over2}(1+\bar a-\bar b),\quad y_*=\bar b+t_*={1\over2}(1-\bar a+\bar b)\ ,$$ and in particular furnishes the maximal value $$f(x_*,y_*)=t_*^2={1\over4}(1-\bar a-\bar b)^2\ .$$
