To tackle the problem of finding the maximum value of \(\frac{A(R) + A(S)}{A(T)}\) where \(T\) is an acute-angled triangle and \(R\) and \(S\) are rectangles inscribed within it, we need to delve into some geometric principles and properties of triangles and rectangles.

Understanding the Areas Involved

First, let's clarify the components of our problem:

• T: An acute-angled triangle ABC.
• R and S: Rectangles inscribed within triangle T, one above the other.
• A(X): The area of polygon X.

The area of triangle \(T\) can be calculated using the formula:

A(T) = \frac{1}{2} \times \text{base} \times \text{height}

Inscribing Rectangles

When we inscribe rectangles \(R\) and \(S\) within triangle \(T\), we need to consider how these rectangles are positioned. Typically, the largest rectangle that can be inscribed in a triangle will have its base along one side of the triangle and its top vertices touching the opposite vertex or sides. The area of each rectangle can be maximized by ensuring that they are as wide as possible while still fitting within the triangle.

Maximizing Areas of Rectangles

Let’s denote the area of rectangle \(R\) as \(A(R)\) and rectangle \(S\) as \(A(S)\). The areas of these rectangles depend on their respective heights and bases:

A(R) = \text{base}_R \times \text{height}_R

A(S) = \text{base}_S \times \text{height}_S

To maximize the combined area of the rectangles relative to the area of the triangle, we can use the properties of similar triangles. If we consider the height of the rectangles as a fraction of the height of triangle \(T\), we can express the areas of the rectangles in terms of the triangle's dimensions.

Finding the Maximum Ratio

The ratio we want to maximize is:

R = \frac{A(R) + A(S)}{A(T)}

Assuming both rectangles are inscribed optimally, we can use the fact that the maximum area of a rectangle inscribed in a triangle is known to be half the area of the triangle. Therefore, if both rectangles are optimally inscribed, we can express:

A(R) + A(S) ≤ A(T)

Thus, the maximum value of the ratio becomes:

R ≤ \frac{A(T)}{A(T)} = 1

Conclusion on the Maximum Value

In summary, the maximum value of \(\frac{A(R) + A(S)}{A(T)}\) is 1, which occurs when the combined area of the rectangles equals the area of the triangle. This situation is theoretically achievable under optimal conditions where the rectangles are perfectly inscribed to utilize the triangle's area fully.

Therefore, the answer to your question is:

Maximum value of \(\frac{A(R) + A(S)}{A(T)}\) is 1.