To find the largest area of the rectangle described in the problem, let's work through the details step-by-step:

Step 1: Define the Area of the Rectangle

Given that the rectangle has:

• Length of the base $x$
• Height $y$

The area $A$ of the rectangle is: $A = x \cdot y$

Step 2: Substitute $y$ in Terms of $x$

We know that the point $(x, y)$ lies on the parabola given by: $y = 64 - x^2$

Substituting this into the area formula, we get: $A = x \cdot (64 - x^2)$ $A = 64x - x^3$

Step 3: Find the Maximum Area

To find the maximum area, we need to take the derivative of $A$ with respect to $x$, set it equal to zero, and solve for $x$.

Derivative of $A$:

$\frac{dA}{dx} = 64 - 3x^2$

Set the Derivative Equal to Zero:

$64 - 3x^2 = 0$ $3x^2 = 64$ $x^2 = \frac{64}{3}$ $x = \sqrt{\frac{64}{3}} = \frac{8}{\sqrt{3}} = \frac{8\sqrt{3}}{3}$

Step 4: Calculate $y$ and $A$

Now, substitute $x = \frac{8\sqrt{3}}{3}$ back into $y = 64 - x^2$ to find $y$: $y = 64 - \left(\frac{8\sqrt{3}}{3}\right)^2$ $y = 64 - \frac{64}{3} = \frac{192 - 64}{3} = \frac{128}{3}$

Now, compute the area $A$: $A = x \cdot y = \frac{8\sqrt{3}}{3} \cdot \frac{128}{3} = \frac{1024\sqrt{3}}{9}$

Final Answer:

$A_{\text{max}} = \frac{1024\sqrt{3}}{9} \text{ square units}$

Would you like more details or have any questions?

Here are five related questions to expand on this topic:

1. How does changing the parabola’s equation affect the maximum area of the rectangle?
2. What happens to the rectangle's area if its base is extended along both sides of the origin?
3. How would you set up a similar optimization problem for an ellipse instead of a parabola?
4. What are some real-world applications of finding the maximum area of a rectangle under a curve?
5. How can you solve this problem using numerical methods instead of calculus?

Tip: Optimization problems often involve substituting one variable in terms of another to simplify the objective function before differentiating.