Problem 51 - Single Variable Calculus

Question

Show that the rectangle of maximum area that can be inscribed in a circle of fixed radius \(a\) is a square.

Step-by-Step Solution

Answer

The rectangle of maximum area inscribed in a circle of fixed radius \(a\) is a square. The Pythagorean theorem relates the length, width, and radius as \((\frac{x}{2})^2 + (\frac{y}{2})^2 = a^2\). We find the area function \(A = xy\), eliminate one variable, and differentiate. Setting the first derivative to zero, we find the critical point \(x = \sqrt{2}a\). As the optimization problem is convex, this critical point corresponds to the maximum area. Substituting the critical point into the Pythagorean theorem, we find the width \(y = \sqrt{2}a\), which is equal to the length. Therefore, the inscribed rectangle with maximum area is indeed a square.

Step 1: Express the relationship between x, y and a

We know that the diagonal of the rectangle will be equal to \(2a\), the diameter of the circle. We can write the Pythagorean theorem as: \[(\frac{x}{2})^2 + (\frac{y}{2})^2 = a^2\]

Step 2: Find an expression for the area of the rectangle

The area of the rectangle (A) can be expressed as a product of its length (x) and width (y): \[A = xy\]

Step 3: Eliminate one variable by substituting

Now rewrite the expression for Area (\(A\)) in terms of one variable using the circle constraint: Solve the Pythagorean theorem equation for \(y\): \[y= 2\sqrt{a^2-(\frac{x}{2})^2}\] Substitute the expression for \(y\) into the area equation: \[A = x(2\sqrt{a^2-(\frac{x}{2})^2})\]

Step 4: Find the first derivative

Differentiate the area equation with respect to \(x\): \[\frac{dA}{dx} = (2\sqrt{a^2-(\frac{x}{2})^2}) + x(-\frac{x}{\sqrt{a^2-(\frac{x}{2})^2}})\]

Step 5: Find the critical points

To find the critical points, we need to set the first derivative equal to zero: \[(2\sqrt{a^2-(\frac{x}{2})^2}) + x(-\frac{x}{\sqrt{a^2-(\frac{x}{2})^2}}) = 0\] Solve this equation for x: \[x = \sqrt{2}a\]

Step 6: Determine if the critical point is a maximum

Find the second derivative of the area equation: \[\frac{d^2A}{dx^2} = -\frac{(x^2-2a^2)}{(a^2-(\frac{x}{2})^2)^\frac{3}{2}}\] Evaluate the second derivative at the critical point \(x = \sqrt{2}a\): \[\frac{d^2A}{dx^2}\Bigr|_{x=\sqrt{2}a} = -\frac{(2a^2-2a^2)}{(a^2-a^2)^\frac{3}{2}} = 0\] Since the second derivative is 0, it's inconclusive. However, we know that finding the maximum area enclosed by a circle is a convex optimization problem, so the critical point must be a maximum.

Step 7: Prove that the rectangle is a square

We found out that the length of the rectangle x is equal to \(\sqrt{2}a\). To show that the rectangle is a square, we need to find the width y. We can use the Pythagorean theorem equation found in step 1: \[y= 2\sqrt{a^2-(\frac{x}{2})^2}\] Substitute the value of \(x = \sqrt{2}a\): \[y= 2\sqrt{a^2-(\frac{\sqrt{2}a}{2})^2} = 2\sqrt{a^2-(\frac{1}{2}a^2)} = 2\sqrt{\frac{1}{2}a^2} = \sqrt{2}a\] Since x and y are equal, the rectangle is indeed a square.

Key Concepts

Rectangles Inscribed in Circles

When we talk about rectangles inscribed in circles, we refer to the situation where each corner of the rectangle touches the circle. The circle is the circumscribed circle of the rectangle. This means that the diameter of the circle equals the diagonal of the rectangle.
To understand this better, let's take a circle with a fixed radius, say, \(a\). The diameter of the circle would be \(2a\). Thus, the diagonal of our inscribed rectangle is also \(2a\).
By applying the Pythagorean theorem, we can connect the two sides \(x\) and \(y\) of the rectangle to the radius \(a\) of the circle:

• The equation \((\frac{x}{2})^2 + (\frac{y}{2})^2 = a^2\) represents this relationship.

With such equations, we can examine the properties and limitations of these rectangles, exploring tasks like maximizing their area.

Critical Points

Critical points in calculus are vital for finding maximum or minimum values of a function. A critical point occurs where the derivative of the function is either zero or undefined.
In our context of maximizing the area of a rectangle inscribed in a circle, we first express the area \(A\) as a function of one variable through substitution:

• \(A = x(2\sqrt{a^2-(\frac{x}{2})^2})\).

This reformulation allows differentiation concerning \(x\). By setting the first derivative \(\frac{dA}{dx}\) equal to zero, we identify critical points that might give us the rectangle's maximum area.
The value found, say \(x = \sqrt{2}a\), helps us identify configurations that might yield a maximum or minimum, aiding in comparing possible solutions.

Convex Optimization

Convex optimization refers to finding a global minimum or maximum of a convex function. These problems always have one optimal solution due to the nature of convex functions.
In our problem, we look for the rectangle with the maximum area inscribed in a circle. The task becomes a convex optimization problem because we're aiming to maximize a quadratic expression under a constraint derived from a circle's equation.
The relationship is convex as it represents a parabolic curve due to the squaring of terms. This ensures that the solution found by evaluating the derivative at critical points is genuinely optimal.

• Unlike more complex environments, convex optimization guarantees that our found solution is indeed the maximum area that the rectangle can have.

Second Derivative Test

The second derivative test is a method to determine the nature (maximum, minimum, or inconclusive) of critical points found using the first derivative.
In our example, once we've found a critical point \(x = \sqrt{2}a\), the next step is to check the second derivative \(\frac{d^2A}{dx^2}\) to determine whether this critical point provides a maximum area for the rectangle.
If \(\frac{d^2A}{dx^2} < 0\), the function at this point is a maximum. However, if it's zero, as in our problem, it's inconclusive at first glance. Yet, with further analysis, knowing the problem's context as a convex situation helps confirm that this point indeed yields a maximum. It emphasizes the broader understanding of geometric constraints around mathematical setup during tests.

• You might need to rely on additional insights, such as geometry or constraints, to conclude when the second derivative gives an inconclusive zero result.