We are tasked with finding the dimensions of the rectangle with the maximum area when its base lies on the $x$-axis and its vertices are above the $x$-axis on the parabola $y = 9 - x^2$.

Step 1: Define the Problem

Let the coordinates of the top corners of the rectangle be $(x, y)$ and $(-x, y)$. Since the rectangle has its base on the $x$-axis, the height of the rectangle is $y$ (the vertical distance from the $x$-axis to the curve). The width of the rectangle is $2x$ (since the base extends from $-x$ to $x$).

From the equation of the parabola $y = 9 - x^2$, we know that the height of the rectangle at any point $x$ is $y = 9 - x^2$.

Thus, the area $A$ of the rectangle as a function of $x$ is given by:

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

So, the area function is:

$A(x) = 2x(9 - x^2) = 18x - 2x^3$

Step 2: Maximize the Area

To maximize the area, we first take the derivative of $A(x)$ with respect to $x$ and set it equal to zero to find the critical points.

$\frac{dA(x)}{dx} = 18 - 6x^2$

Set the derivative equal to zero:

$18 - 6x^2 = 0$

Solve for $x^2$:

$6x^2 = 18$

$x^2 = 3$

$x = \pm\sqrt{3} \approx \pm 1.732$

Since $x$ represents half the width of the rectangle and must be positive, we take $x = \sqrt{3} \approx 1.732$.

Step 3: Find the Maximum Area

To ensure that this critical point corresponds to a maximum, we check the second derivative of $A(x)$:

$\frac{d^2A(x)}{dx^2} = -12x$

At $x = \sqrt{3}$:

$\frac{d^2A(x)}{dx^2} = -12(\sqrt{3}) \approx -20.784$

Since the second derivative is negative, this indicates that $x = \sqrt{3}$ is a maximum.

Step 4: Calculate the Dimensions and Area

Now, we calculate the dimensions and the area:

• The height of the rectangle is $y = 9 - x^2 = 9 - 3 = 6$.
• The width of the rectangle is $2x = 2 \times \sqrt{3} \approx 3.464$.

Thus, the shorter dimension (height) is approximately 6, and the longer dimension (width) is approximately 3.46.

Finally, the maximum area is:

$A_{\text{max}} = 2x \times (9 - x^2) = 2 \times \sqrt{3} \times 6 \approx 6 \times 3.464 \approx 20.784$

Final Answer:

• Shorter dimension (height) = 6.
• Longer dimension (width) = 3.46.
• Maximum area = 20.78.

Would you like any further explanation or assistance? Here are some related questions for practice:

1. How do you determine whether a critical point is a maximum or minimum?
2. How do you find the area of a rectangle inscribed in a different type of curve, like a circle?
3. Can you use this same method for finding maximum areas of other shapes?
4. What happens if you change the equation of the parabola? How would that affect the dimensions of the rectangle?
5. What is the relationship between the first and second derivatives in optimization problems?

Tip: Always check the second derivative to confirm whether a critical point gives a maximum or minimum.