Question As shown in Figure 1, two rectangles $R$ and $S$ are placed inside an acute triangle $T$. Let $A(X)$ denote the area of polygon $X$. Find the maximum value of $\frac{A(R)+A(S)}{A(T)}$ or prove that the maximum value does not exist.
As shown in Figure 1, let the heights of rectangles $R$ and $S$ be $a$ and $b$, respectively, the height of the triangle be $h$, and the distance from the vertex opposite the base to rectangle $S$ be $c$. Then $a+b+c=h$.
In reference [1], the ratio of areas is expressed using similar triangles as
$$
\frac{A(R)+A(S)}{A(T)}=\frac{2}{h^{2}}(a b+b c+a c),
$$

Then, the local adjustment method is used, first fixing the value of $a$.
Thus, $a b+b c+a c=a(h-a)+b c$, and $b+c$ is a constant. When $b=c$, $b c$ reaches its maximum value.

Therefore, the problem is reduced to finding the maximum value of $2 a b+b^{2}$ under the condition $a+2 b=h$.

Substituting $a=h-2 b$ yields the quadratic function in terms of $b$: $-3 b^{2}+2 b h$.

By completing the square, it is known that when $a=b=c=\frac{h}{3}$, $\frac{A(R)+A(S)}{A(T)}$ reaches its maximum value of $\frac{2}{3}$.