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We are given the equation of the parabola \( h(x) = -2x^2 + 6x \), and we need to find the dimensions and the area of the largest shaded right-angled triangle, which is inscribed between the curve, the positive x-axis, and the y-axis.

### Step 1: Set up the general form of the triangle's area
The vertices of the triangle are at points:
- The origin \( (0, 0) \),
- A point on the x-axis \( (x, 0) \),
- A point on the curve \( (x, h(x)) \).

The area \( A \) of this right triangle can be written as:
\[
A = \frac{1}{2} \times \text{base} \times \text{height}
\]
where the base is \( x \) (the distance along the x-axis) and the height is \( h(x) \) (the distance from the x-axis to the curve).

Thus, the area function becomes:
\[
A(x) = \frac{1}{2} x \cdot h(x)
\]
Substitute \( h(x) = -2x^2 + 6x \) into the area equation:
\[
A(x) = \frac{1}{2} x (-2x^2 + 6x)
\]
Simplifying:
\[
A(x) = \frac{1}{2} (-2x^3 + 6x^2) = -x^3 + 3x^2
\]

### Step 2: Maximize the area
To find the value of \( x \) that maximizes the area, we take the derivative of \( A(x) \) with respect to \( x \) and set it equal to zero:
\[
\frac{dA}{dx} = -3x^2 + 6x
\]
Setting this equal to zero:
\[
-3x^2 + 6x = 0
\]
Factor the equation:
\[
-3x(x - 2) = 0
\]
Thus, \( x = 0 \) or \( x = 2 \).

Since \( x = 0 \) corresponds to no triangle (area = 0), the value of \( x \) that maximizes the area is \( x = 2 \).

### Step 3: Calculate the maximum area
Substitute \( x = 2 \) back into the equation for \( A(x) \):
\[
A(2) = - (2)^3 + 3(2)^2 = -8 + 12 = 4
\]

Thus, the maximum area of the triangle is 4 square units.

### Step 4: Find the dimensions of the triangle
- The base of the triangle is \( x = 2 \).
- The height is \( h(2) = -2(2)^2 + 6(2) = -8 + 12 = 4 \).

Therefore, the dimensions of the largest triangle are:
- Base: 2 units
- Height: 4 units

### Final Answer:
- **Dimensions:** Base = 2 units, Height = 4 units
- **Maximum Area:** 4 square units

Would you like more details or have any questions?

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**Relative Questions:**
1. How do you derive the critical points in maximizing a function?
2. What role does the derivative play in finding the maximum area?
3. Can we apply this method to other curves besides parabolas?
4. How do we determine if a critical point is a maximum or a minimum?
5. What would happen if the curve's equation were different?

**Tip:** When maximizing the area of a geometric figure, always set up an equation for the area and then use calculus to find critical points where the area is maximized.

Sketched below is the graph of h(x) = -2x^2 + 6x. What are the dimensions and area of the largest shaded right-angled triangle with maximum area, that can be drawn from the origin, between h, the positive x-axis and the positive y-axis?

Learn how to find the maximum area of a right-angled triangle inscribed under the curve h(x) = -2x^2 + 6x. This problem involves optimization using calculus and the properties of quadratic functions. Ideal for Grades 11-12 students.

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