Question

Find the dimensions of the rectangle of greatest area that can be inscribed in the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1\).

Short answer

The dimensions are \(\frac{a}{\sqrt{2}}\) by \(\frac{b}{\sqrt{2}}\).

Step-by-step solution

Understand the Problem

We need to find the dimensions of the rectangle of greatest possible area that can be inscribed in the given ellipse. We define the coordinates of the rectangle's vertices as \((x, y), (-x, y), (x, -y), (-x, -y)\) which satisfy the ellipse equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). The rectangle is symmetrical with respect to both axes.

Derive the Area Formula

The area of the rectangle can be expressed as \(A = 2x \cdot 2y = 4xy\). Since \(x\) and \(y\) are on the ellipse, they need to satisfy the ellipse equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Thus, we can express the area as a function of one variable only using this constraint.

Substitute Using the Ellipse Equation

From the ellipse equation, express \(y\) in terms of \(x\):\[y = b\sqrt{1 - \frac{x^2}{a^2}}\]Substitute this expression into the area formula:\[A(x) = 4x \cdot b\sqrt{1 - \frac{x^2}{a^2}}\]

Differentiate the Area Function

Find the derivative of the area function \(A(x) = 4xb\sqrt{1 - \frac{x^2}{a^2}}\):First, set up the derivative using the chain rule:\[A'(x) = 4b\left( \sqrt{1 - \frac{x^2}{a^2}} - \frac{x^2}{a^2\sqrt{1 - \frac{x^2}{a^2}}} \right)\]

Set Derivative to Zero

To find the critical points, solve the equation \(A'(x) = 0\):\[\sqrt{1 - \frac{x^2}{a^2}} = \frac{x^2}{a^2\sqrt{1 - \frac{x^2}{a^2}}}\]This simplifies to \(x^2 = a^2 - x^2\), leading to \(2x^2 = a^2\), thus \(x^2 = \frac{a^2}{2}\). Solve for \(x\):\[x = \frac{a}{\sqrt{2}}\]

Find Corresponding y-value

Substitute \(x = \frac{a}{\sqrt{2}}\) back into the ellipse equation to find \(y\):\[\frac{a^2/2}{a^2} + \frac{y^2}{b^2} = 1\]\[\frac{1}{2} + \frac{y^2}{b^2} = 1\]\[\frac{y^2}{b^2} = \frac{1}{2}\]Solving this gives \(y = \frac{b}{\sqrt{2}}\).

Conclude the Dimensions

The dimensions of the rectangle of greatest area inscribed in the ellipse are \(\frac{a}{\sqrt{2}}\) by \(\frac{b}{\sqrt{2}}\).

Key concepts

Rectangle Inscribed in Ellipse

A rectangle is inscribed in an ellipse when each of its vertices touches the edge of the ellipse. Visualize the rectangle's vertices as points that move along the ellipse's boundary. The shape of the ellipse constrains the possible positions of these points.

The rectangle's symmetry makes it simpler to analyze. By focusing on one quadrant of the ellipse, we can define a pair of vertices as overlapping two axes, specifically:

• Top right corner at defined as \((x, y)\).
• Other vertices located at \((-x, y)\), \((x, -y)\), and \((-x, -y)\).

Understanding this setup is crucial for modeling the problem mathematically.

Ellipse Equation

The equation of an ellipse centered at the origin is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). This equation describes all the possible \((x, y)\) points that lie on the ellipse.

Here, \(a\) and \(b\) represent the ellipse's semi-major and semi-minor axes, respectively. They define the ellipse's width and height.

• The equation is crucial for finding \(y\) in terms of \(x\)
• Setting the stage for calculating things like area or distances within the ellipse.

Plugging any point \((x, y)\) into the ellipse equation confirms if it truly lies on the ellipse—a foundational step to solving the problem.

Derivative and Critical Points

Calculus involves finding the derivative of functions to identify their critical points.

To maximize the area of the rectangle, we first represent the area \(A(x)\) as a function of the rectangle's dimensions and then take its derivative. This involves substituting \(y\) expressed in terms of \(x\), derived from the ellipse equation, into the area formula\[A(x) = 4x \, b \sqrt{1 - \frac{x^2}{a^2}}\].

• Differentiating this function helps find where the area reaches its peak potential, or its critical points.
• Critical points occur where the derivative is zero or undefined.

Solving \(A'(x) = 0\) indicates these critical points, crucial for pinpointing optimal dimensions.

Maximizing Area

Maximizing the area requires us to solve for \(x\) using the condition \(A'(x) = 0\). This process identifies the dimensions where the rectangle's area is largest.

After deriving \(A(x)\) and finding the critical point by solving the derivative \(A'(x)\), we calculate the corresponding \(y\) based on the ellipse equation:

• This results in \(x = \frac{a}{\sqrt{2}}\).
• Substitute \(x\) into the ellipse equation to find \(y = \frac{b}{\sqrt{2}}\).

The final result shows that the rectangle's maximum inscribed area has dimensions \(\frac{a}{\sqrt{2}}\) by \(\frac{b}{\sqrt{2}}\). This method highlights the beauty of optimization in calculus.

Geometry and Calculus

The interplay between geometry and calculus is evident in this problem. Geometry provides the context with shapes and their relationships, while calculus offers tools to analyze and optimize those shapes.

This task emphasizes how:

• Geometric constraints like the ellipse direct the possible dimensions of the rectangle.
• Calculus identifies the specific points where the rectangle's area is maximized.

By merging these fields, we solve practical problems involving optimization, a valuable skill in mathematics and beyond.