Let's walk through solving the problem step-by-step based on the image:

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.