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The problem asks to find the maximum area \( A_{\text{max}} \) of a rectangle that is inscribed in a right triangle, where the triangle has legs of length \( a = 22 \, \text{cm} \). The sides of the rectangle have lengths \( x \) and \( y \), with the base of the triangle forming the hypotenuse.

### Step-by-step Explanation:

1. **Geometry of the Triangle**:
   - The triangle is a right triangle with both legs equal to \( a = 22 \, \text{cm} \).
   - The equation of the hypotenuse can be derived as:
     \[
     y = -\frac{22}{22}x + 22 \quad \Rightarrow \quad y = 22 - x
     \]
     This gives the relationship between \( x \) and \( y \) as \( y = 22 - x \).

2. **Area of the Rectangle**:
   - The area \( A \) of the rectangle is given by \( A = x \times y \).
   - Substituting the equation of the hypotenuse, we get:
     \[
     A(x) = x \times (22 - x) = 22x - x^2
     \]
     This is the area function \( A(x) \) to maximize.

3. **Maximization**:
   - To find the maximum area, we take the derivative of \( A(x) \) with respect to \( x \):
     \[
     \frac{dA}{dx} = 22 - 2x
     \]
     Setting the derivative equal to zero to find the critical points:
     \[
     22 - 2x = 0 \quad \Rightarrow \quad x = 11 \, \text{cm}
     \]
   - To confirm that this is a maximum, we check the second derivative:
     \[
     \frac{d^2A}{dx^2} = -2
     \]
     Since the second derivative is negative, \( x = 11 \, \text{cm} \) gives a maximum.

4. **Maximum Area**:
   - Substituting \( x = 11 \) into the equation for \( y \):
     \[
     y = 22 - 11 = 11 \, \text{cm}
     \]
   - The maximum area is:
     \[
     A_{\text{max}} = 11 \times 11 = 121 \, \text{cm}^2
     \]

Thus, the maximum area of the rectangle is \( \boxed{121 \, \text{cm}^2} \).

Let me know if you'd like further details or have any questions! 

Here are 5 related questions to consider:
1. How would the area change if one leg of the triangle were longer than the other?
2. What if the triangle were an isosceles right triangle but with different leg lengths?
3. Can you express the result in terms of a general leg length \( a \)?
4. How would this problem change if the rectangle were inscribed in a different shape, like a circle?
5. What mathematical techniques are useful for maximizing areas or volumes in general?

**Tip**: Always check whether the second derivative is negative when trying to confirm that you have found a maximum in a maximization problem!

Given a right triangle with legs each of length 22 cm, find the maximum area of a rectangle inscribed in this triangle, where the rectangle's sides are along the legs of the triangle.

Discover how to maximize the area of a rectangle inscribed in a right triangle with legs of 22 cm each. This solution uses calculus to find the optimal dimensions of the rectangle that yield the largest area, demonstrating the application of derivative tests for optimization.

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