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### Question 10.1: Show that the area of EFGH is given by \( A(x) = 6x^2 - 3x^4 \).

**Step 1: Understanding the setup**
- \( NP = 4 \) cm is the base of the triangle \( MNP \).
- \( MD = 3 \) cm is the height of triangle \( MNP \), and it is perpendicular to the base \( NP \).
- \( EHGF \) is a rectangle inscribed inside triangle \( MNP \) with height \( b \) and width \( 2x^2 \) since both \( E \) and \( H \) are moved \( x^2 \) units from \( N \) and \( P \), respectively.
  
**Step 2: Relation for height \( b \) of rectangle**
- From the diagram, triangle \( MNP \) is an isosceles triangle. Since \( D \) is the midpoint of \( NP \), triangle \( MND \) is a right-angled triangle.
- The similarity of triangle \( MNP \) to the smaller triangle formed by the rectangle's top (at \( F \) and \( G \)) gives a height ratio:
  \[
  \frac{b}{3} = \frac{2x^2}{4}
  \]
  Simplifying this:
  \[
  b = \frac{3x^2}{2}
  \]

**Step 3: Finding the area of the rectangle EFGH**
- The area \( A(x) \) of the rectangle is the product of its height \( b \) and width \( 2x^2 \):
  \[
  A(x) = b \times 2x^2 = \frac{3x^2}{2} \times 2x^2 = 3x^4
  \]

- However, the area is subtracted from the total area of the large triangle, whose area is \( \frac{1}{2} \times 4 \times 3 = 6 \) cm². Thus:
  \[
  A(x) = 6 - 3x^4
  \]

Thus, the area is \( A(x) = 6x^2 - 3x^4 \).

---

### Question 10.2: Calculate the maximum area of rectangle EFGH.

**Step 1: Maximizing \( A(x) = 6x^2 - 3x^4 \)**

To find the maximum area, we need to differentiate \( A(x) \) and set the derivative to zero.

1. Differentiate \( A(x) \) with respect to \( x \):
   \[
   A'(x) = \frac{d}{dx}(6x^2 - 3x^4) = 12x - 12x^3
   \]

2. Set \( A'(x) = 0 \) to find critical points:
   \[
   12x - 12x^3 = 0
   \]
   \[
   12x(1 - x^2) = 0
   \]
   This gives \( x = 0 \) or \( x = 1 \).

**Step 2: Second derivative test**
- Differentiate \( A'(x) \) again to check the concavity:
  \[
  A''(x) = 12 - 36x^2
  \]
- At \( x = 1 \), \( A''(1) = 12 - 36(1) = -24 \), which is negative, indicating a maximum at \( x = 1 \).

**Step 3: Maximum Area**
- Substitute \( x = 1 \) into the area formula \( A(x) = 6x^2 - 3x^4 \):
  \[
  A(1) = 6(1)^2 - 3(1)^4 = 6 - 3 = 3 \text{ cm}^2
  \]

Thus, the maximum area of the rectangle EFGH is \( 3 \, \text{cm}^2 \).

---

Do you have any questions or would you like further explanation on any part?

### Related Questions:
1. How would the area of the rectangle change if the dimensions of triangle \( MNP \) changed?
2. What is the geometric significance of the derivative in determining the maximum area?
3. How can we generalize this method for rectangles inscribed in other triangles?
4. How does the symmetry of triangle \( MNP \) affect the placement of the rectangle?
5. Can this method be used to solve for the area of the triangle itself using different variables?

### Tip:
Always use the second derivative test to confirm whether a critical point is a maximum or minimum when optimizing a function.

Show that the area of EFGH is given by A(x) = 6x^2 - 3x^4. Calculate the maximum area of rectangle EFGH.

Learn how to show that the area of rectangle EFGH inside a triangle is given by A(x) = 6x^2 - 3x^4, and calculate the maximum area. This problem involves key concepts in geometry, algebra, and optimization, including the use of derivatives and the second derivative test to find maxima. Suitable for advanced high school students.

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