solve step by step Q2 Find the dimensions of a rectangle of largest area that
has its base on the x − axisand its other two sides above
thex − axis and lying on the parabola y = 8 − x
2
. [7]]

To solve this problem, we need to find the dimensions of a rectangle with the largest area, where the base of the rectangle is on the \( x \)-axis, and its upper corners lie on the parabola \( y = 8 - x^2 \).

### Step-by-Step Solution:

1. **Equation of the Parabola:**
   The equation of the parabola is given as:
   \[
   y = 8 - x^2
   \]
   This describes the shape where the rectangle's top two corners will lie.

2. **Coordinates of the Rectangle:**
   Since the base is on the \( x \)-axis, and the other two sides lie on the parabola, we can assume that the rectangle is symmetric about the \( y \)-axis.
   
   Let the top right corner of the rectangle have coordinates \( (x, y) \). By symmetry, the top left corner will be at \( (-x, y) \). The base of the rectangle will extend from \( -x \) to \( x \), so the width of the rectangle is \( 2x \).

3. **Height of the Rectangle:**
   The height of the rectangle is the \( y \)-coordinate of the top corners, which is given by the equation of the parabola:
   \[
   y = 8 - x^2
   \]
   
4. **Area of the Rectangle:**
   The area \( A \) of the rectangle is given by the product of its width and height. Thus:
   \[
   A(x) = \text{width} \times \text{height} = 2x \times (8 - x^2)
   \]
   Simplifying this expression:
   \[
   A(x) = 2x(8 - x^2) = 16x - 2x^3
   \]
   
5. **Maximizing the Area:**
   To find the value of \( x \) that maximizes the area, we need to take the derivative of \( A(x) \) with respect to \( x \), and then set the derivative equal to zero to find critical points.

   First, differentiate \( A(x) \):
   \[
   A'(x) = \frac{d}{dx}(16x - 2x^3) = 16 - 6x^2
   \]
   
   Now, set \( A'(x) = 0 \) to find the critical points:
   \[
   16 - 6x^2 = 0
   \]
   Solving for \( x \):
   \[
   6x^2 = 16
   \]
   \[
   x^2 = \frac{16}{6} = \frac{8}{3}
   \]
   \[
   x = \pm \sqrt{\frac{8}{3}} = \pm \frac{2\sqrt{6}}{3}
   \]
   Since \( x \) represents half the width of the rectangle, we only consider the positive value:
   \[
   x = \frac{2\sqrt{6}}{3}
   \]
   
6. **Second Derivative Test:**
   To confirm that this value of \( x \) gives a maximum, we use the second derivative test. The second derivative of \( A(x) \) is:
   \[
   A''(x) = \frac{d}{dx}(16 - 6x^2) = -12x
   \]
   Substituting \( x = \frac{2\sqrt{6}}{3} \) into \( A''(x) \):
   \[
   A''\left(\frac{2\sqrt{6}}{3}\right) = -12 \times \frac{2\sqrt{6}}{3} = -8\sqrt{6}
   \]
   Since \( A''(x) < 0 \), this confirms that \( x = \frac{2\sqrt{6}}{3} \) is a maximum.

7. **Height and Dimensions of the Rectangle:**
   The height of the rectangle is:
   \[
   y = 8 - x^2 = 8 - \frac{8}{3} = \frac{24}{3} - \frac{8}{3} = \frac{16}{3}
   \]
   
   Therefore, the dimensions of the rectangle are:
   - Width: \( 2x = 2 \times \frac{2\sqrt{6}}{3} = \frac{4\sqrt{6}}{3} \)
   - Height: \( \frac{16}{3} \)

### Final Answer:
The dimensions of the rectangle with the largest area are:
- Width: \( \frac{4\sqrt{6}}{3} \)
- Height: \( \frac{16}{3} \)

Would you like further clarification on any steps?

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### 5 Relative Questions:
1. What is the importance of using the second derivative test in optimization problems?
2. How would the result change if the parabola had a different equation, such as \( y = 10 - x^2 \)?
3. Can you explain how the symmetry of the rectangle simplifies the problem-solving process?
4. How can we apply this method to find the largest rectangle under other curves?
5. What would happen if we allowed the rectangle to extend beyond the \( x \)-axis?

### Tip:
When maximizing or minimizing areas, always confirm whether the critical points are maxima or minima using the second derivative test.

Find the dimensions of a rectangle of largest area that has its base on the x-axis and its other two sides above the x-axis, lying on the parabola y = 8 - x^2.

Learn how to find the dimensions of the rectangle with the largest area under the parabola y = 8 - x^2. This problem demonstrates key calculus concepts like optimization and the second derivative test. Suitable for students in Grades 11-12.

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