Let's go through the steps one by one to solve the problem.

[A] Finding the Area as a Function of $x$:

The rectangle is inscribed with its base on the $x$-axis and its upper corners on the parabola $y = 3600 - x^2$.

Let the coordinates of the upper corners of the rectangle be $(x, y)$ and $(-x, y)$. Since the base is on the $x$-axis, the width of the rectangle is $2x$, and the height is $y$.

Given $y = 3600 - x^2$, the area $A(x)$ of the rectangle is: $A(x) = \text{width} \times \text{height} = 2x \times y = 2x(3600 - x^2)$ So, the area function is: $A(x) = 2x(3600 - x^2) = 7200x - 2x^3$

[B] Finding the Domain of $A(x)$:

The domain is determined by the values of $x$ for which the rectangle is in the first quadrant.

Since the rectangle's upper corners are on the parabola $y = 3600 - x^2$, $y$ must be non-negative: $3600 - x^2 \geq 0$ $x^2 \leq 3600$ $|x| \leq 60$ Thus, $x$ is in the interval $[-60, 60]$.

However, since the rectangle is in the first quadrant, $x$ must be non-negative. Therefore, the domain is: $\text{Domain: } [0, 60]$

[C] Finding the Derivative $A'(x)$:

To find the critical points, we differentiate $A(x)$ with respect to $x$: $A(x) = 7200x - 2x^3$ $A'(x) = 7200 - 6x^2$

[D] Finding the Critical Value(s):

To find the critical values, set $A'(x) = 0$: $7200 - 6x^2 = 0$ $6x^2 = 7200$ $x^2 = 1200$ $x = \sqrt{1200} = 20\sqrt{3}$

Since the critical value $x = 20\sqrt{3}$ lies within the domain $[0, 60]$, it is valid.

[E] Determining Where $A(x)$ is Increasing and Decreasing:

To determine the intervals where $A(x)$ is increasing or decreasing, analyze the sign of $A'(x)$: $A'(x) = 7200 - 6x^2$

• $A'(x) > 0$ when $x^2 < 1200$, i.e., $x \in [0, 20\sqrt{3})$.
• $A'(x) < 0$ when $x^2 > 1200$, i.e., $x \in (20\sqrt{3}, 60]$.

So: $\text{Increasing: } [0, 20\sqrt{3})$ $\text{Decreasing: } (20\sqrt{3}, 60]$

[F] Conclusion Regarding the Absolute Extrema:

Since $A(x)$ goes from increasing to decreasing at $x = 20\sqrt{3}$, by the first derivative test, $A(x)$ has an absolute maximum at $x = 20\sqrt{3}$.

• Conclusion: Since $A(x)$ goes from increasing to decreasing, by the first derivative test, we have an absolute maximum at the critical value.

[G] Maximum Area of the Rectangle:

To find the maximum area, substitute $x = 20\sqrt{3}$ into the area function $A(x)$: $A(20\sqrt{3}) = 7200(20\sqrt{3}) - 2(20\sqrt{3})^3$ $A(20\sqrt{3}) = 144000\sqrt{3} - 2(8000\sqrt{3}) = 144000\sqrt{3} - 16000\sqrt{3} = 128000\sqrt{3}$ The maximum area is: $\text{Maximum Area} = 128000\sqrt{3}$

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Do you want more details or have any questions?

Here are 5 related questions for practice:

1. How would the area function change if the parabola were $y = 4000 - x^2$?
2. What would be the domain if the rectangle were allowed to extend into the second quadrant?
3. How does the shape of the parabola affect the maximum area?
4. What happens to the maximum area if the parabola shifts vertically upward?
5. Could this method be used for a parabola that opens horizontally?

Tip: Always double-check your critical points and the sign of the derivative in optimization problems to confirm the nature of the extremum!