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### Problem Summary:
A rectangle is inscribed in a circle with radius 6, and point \( P = (x, y) \) is a vertex of the rectangle on the circle. You are asked to express the area and perimeter of the rectangle as functions of \( x \), and to find for what values of \( x \) these functions are largest.

### Step-by-Step Solution:

#### (a) Express the area \( A \) of the rectangle as a function of \( x \).

The equation of the circle is given by:
\[
x^2 + y^2 = 36 \quad \text{(since the radius of the circle is 6)}
\]
Solving for \( y \):
\[
y = \sqrt{36 - x^2}
\]
The total width of the rectangle is \( 2x \) (since \( P = (x, y) \) is a vertex, and the rectangle is symmetric about the origin). The total height is \( 2y \), so the area \( A \) of the rectangle is:
\[
A(x) = \text{width} \times \text{height} = 2x \times 2y = 4x \sqrt{36 - x^2}
\]

#### (b) Express the perimeter \( p \) of the rectangle as a function of \( x \).

The perimeter is the sum of all the sides of the rectangle. Since the width is \( 2x \) and the height is \( 2y = 2\sqrt{36 - x^2} \), the perimeter is:
\[
p(x) = 2 \times \text{width} + 2 \times \text{height} = 2(2x) + 2(2\sqrt{36 - x^2}) = 4x + 4\sqrt{36 - x^2}
\]

#### (c) Graph \( A(x) \). For what value of \( x \) is \( A \) largest?

The function for the area is \( A(x) = 4x \sqrt{36 - x^2} \). To find when \( A(x) \) is largest, we take the derivative of \( A(x) \) and find the critical points.

Let’s compute the critical value of \( x \).

1. Differentiate \( A(x) \):
\[
A'(x) = 4\sqrt{36 - x^2} + 4x \left(\frac{-2x}{2\sqrt{36 - x^2}}\right)
\]
Simplify:
\[
A'(x) = 4\sqrt{36 - x^2} - \frac{4x^2}{\sqrt{36 - x^2}}
\]
Setting \( A'(x) = 0 \), we solve:
\[
4\sqrt{36 - x^2} = \frac{4x^2}{\sqrt{36 - x^2}}
\]
Multiply both sides by \( \sqrt{36 - x^2} \):
\[
4(36 - x^2) = 4x^2
\]
Simplify:
\[
144 - 4x^2 = 4x^2
\]
\[
144 = 8x^2
\]
\[
x^2 = 18
\]
\[
x = \sqrt{18} = 4.24
\]

So, the area \( A(x) \) is largest when \( x \approx 4.24 \).

#### (d) Graph \( p(x) \). For what value of \( x \) is \( p \) largest?

The function for the perimeter is \( p(x) = 4x + 4\sqrt{36 - x^2} \). To find when \( p(x) \) is largest, we differentiate \( p(x) \).

1. Differentiate \( p(x) \):
\[
p'(x) = 4 - \frac{4x}{\sqrt{36 - x^2}}
\]
Set \( p'(x) = 0 \) to find critical points:
\[
4 = \frac{4x}{\sqrt{36 - x^2}}
\]
Solve:
\[
\sqrt{36 - x^2} = x
\]
Square both sides:
\[
36 - x^2 = x^2
\]
\[
36 = 2x^2
\]
\[
x^2 = 18
\]
\[
x = \sqrt{18} = 4.24
\]

So, the perimeter \( p(x) \) is also largest when \( x \approx 4.24 \).

### Summary of Results:
- The area \( A(x) = 4x \sqrt{36 - x^2} \) is largest when \( x \approx 4.24 \).
- The perimeter \( p(x) = 4x + 4\sqrt{36 - x^2} \) is largest when \( x \approx 4.24 \).

Would you like more details or have any other questions?

### Related Questions:
1. What is the significance of the critical points found for \( A(x) \) and \( p(x) \)?
2. Can you describe why the symmetry of the rectangle in the circle leads to the largest values at \( x \approx 4.24 \)?
3. How would you modify the problem if the circle had a different radius, say 10 instead of 6?
4. What happens to the area and perimeter as \( x \) approaches 0 or 6?
5. Could this method be applied to other shapes, such as ellipses, and how would the equations change?

### Tip:
Always check your boundary values when working with maxima and minima, especially for geometric problems like this where the physical constraints (like \( x \) being less than the radius) help guide the solution.

A rectangle is inscribed in a circle of radius 6. Let P = (x, y) be a point on the circle. Express the area and perimeter of the rectangle as functions of x, and find the values of x for which the area and perimeter are largest.

Learn how to express the area and perimeter of a rectangle inscribed in a circle with radius 6 as functions of x. Explore the derivatives to find the values of x where the area and perimeter are largest. Suitable for high school students studying geometry and calculus.

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