We are given a parabola:

$y = 9 - x^2$

A rectangle is constructed with its base on the $x$-axis and its upper two vertices on the parabola. Let’s define its dimensions:

• The base extends symmetrically from $-x$ to $x$, so the width is $2x$.
• The height of the rectangle is determined by the $y$-coordinate of the parabola at $x$, which is $9 - x^2$.

Step 1: Define the Area Function

The area $A$ of the rectangle is given by:

$A(x) = \text{width} \times \text{height} = (2x)(9 - x^2)$

Expanding:

$A(x) = 18x - 2x^3$

Step 2: Find the Maximum Area

To maximize $A(x)$, take the derivative and set it to zero:

$A'(x) = \frac{d}{dx} (18x - 2x^3)$

$A'(x) = 18 - 6x^2$

Setting $A'(x) = 0$:

$18 - 6x^2 = 0$

$6x^2 = 18$

$x^2 = 3$

$x = \sqrt{3} \approx 1.73$

Step 3: Find the Maximum Area

The width:

$\text{Width} = 2x = 2\sqrt{3} \approx 3.46$

The height:

$\text{Height} = 9 - x^2 = 9 - 3 = 6$

Thus, the maximum area is:

$A_{\max} = (2\sqrt{3}) \times 6 = 12\sqrt{3} \approx 20.78$

Final Answer:

• Shorter dimension (height): 6.00
• Longer dimension (width): 3.46
• Maximum area: 20.78

Would you like more details or a graph?

Related Questions:

1. How do we verify that this is a maximum using the second derivative?
2. What happens to the area if the parabola equation changes?
3. Can we solve this problem using Lagrange multipliers?
4. What is the significance of symmetry in maximizing the area?
5. How would the dimensions change if the parabola was $y = 16 - x^2$?

Math Tip:

Always check if a critical point is a maximum or minimum by using the second derivative test!