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.