https://file.aiquickdraw.com/mathbot/file/1729470863557ogtqgnba.jpg

We are tasked with finding the dimensions (height and width) of the rectangle that gives the largest area, based on the given parabola equation \( y = k - x^2 \) with \( k = 8 \).

### Step 1: Defining the Area of the Rectangle
The vertices of the rectangle lie on the parabola \( y = 8 - x^2 \). For a symmetric rectangle about the y-axis, the width of the rectangle is from \( -x \) to \( x \), so the width is \( 2x \).

The height of the rectangle is the y-coordinate of the points on the parabola at \( x \), which is given by:
\[
\text{Height} = y = 8 - x^2
\]

### Step 2: Writing the Area Function
The area \( A \) of the rectangle is the product of its width and height:
\[
A(x) = \text{width} \times \text{height} = 2x \times (8 - x^2)
\]
Simplifying this:
\[
A(x) = 2x(8 - x^2) = 16x - 2x^3
\]

### Step 3: Maximizing the Area
To find the value of \( x \) that maximizes the area, we take the derivative of the area function and set it equal to zero.

\[
\frac{dA}{dx} = 16 - 6x^2
\]
Setting \( \frac{dA}{dx} = 0 \) gives:
\[
16 - 6x^2 = 0 \quad \Rightarrow \quad 6x^2 = 16 \quad \Rightarrow \quad x^2 = \frac{8}{3} \quad \Rightarrow \quad x = \sqrt{\frac{8}{3}} = \frac{2\sqrt{6}}{3}
\]

### Step 4: Calculating the Dimensions
- **Width**: The width is \( 2x \):
  \[
  \text{Width} = 2 \times \frac{2\sqrt{6}}{3} = \frac{4\sqrt{6}}{3} \approx 3.27
  \]

- **Height**: The height is \( 8 - x^2 \):
  \[
  \text{Height} = 8 - \frac{8}{3} = \frac{24}{3} - \frac{8}{3} = \frac{16}{3} \approx 5.33
  \]

### Final Answer:
- Width ≈ 3.27 units
- Height ≈ 5.33 units

Would you like more details on the steps? Let me know if you have any questions.

Here are five related questions to deepen your understanding:
1. How does the symmetry of the parabola affect the width of the rectangle?
2. What would happen to the area if the parabola equation had a different \( k \)-value?
3. How does the second derivative of the area function confirm whether the critical point is a maximum or minimum?
4. What are some alternative ways to maximize the area using different methods (e.g., Lagrange multipliers)?
5. How could we solve this using numerical methods instead of calculus?

**Tip**: To ensure you're maximizing or minimizing a function, always check the second derivative for concavity or use a sign chart for the first derivative!

Find the dimensions that give the largest area for the rectangle shown in the figure. Its base is on the x-axis, and its other two vertices are above the x-axis, lying on the parabola y = k - x^2, where k = 8.

Find the dimensions of a rectangle with the largest area under the parabola y = 8 - x^2. Learn step-by-step how to use calculus to maximize the area by solving for critical points and applying the first derivative test. Suitable for Grade 12 to college-level students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation