Problem 38 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The dimensions of the rectangle are \(\sqrt{2}a \times \sqrt{2}b\).

Step 1: Understand the Problem

We need to find the dimensions of a rectangle that has the maximum possible area while being inscribed in an ellipse given by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). The rectangle will be oriented with its sides parallel to the axes.

Step 2: Set Up Variables and Functions

Let's assume that the vertices of the rectangle are \((x, y), (-x, y), (x, -y), (-x, -y)\). The area of the rectangle is \(A = 2x \, \times \, 2y = 4xy\). We express \(y\) in terms of \(x\) using the ellipse equation: \(y = b \sqrt{1 - \frac{x^2}{a^2}}\). Substitute \(y\) in the area formula to get \(A(x) = 4x b \sqrt{1 - \frac{x^2}{a^2}}\).

Step 3: Maximize the Area Function

We need to find the maximum value of \(A(x)\). First, find the derivative \(A'(x)\) using the product, chain, and power rules. The derivative is \(A'(x) = 4b \left(\sqrt{1 - \frac{x^2}{a^2}} - \frac{x^2}{a^2 \sqrt{1 - \frac{x^2}{a^2}}}\right)\). Set \(A'(x) = 0\) to find the critical points that could give the maximum area.

Step 4: Solve the Critical Point Equation

Set \(4b \left(\sqrt{1 - \frac{x^2}{a^2}} - \frac{x^2}{a^2 \sqrt{1 - \frac{x^2}{a^2}}}\right) = 0\). Simplifying gives \(1 - \frac{2x^2}{a^2} = 0\), which results in \(x^2 = \frac{a^2}{2}\). Therefore, \(x = \frac{a}{\sqrt{2}}\).

Step 5: Find Corresponding y-Value

Substitute \(x = \frac{a}{\sqrt{2}}\) back into the ellipse equation to find \(y\). This gives \(y = b \sqrt{1 - \frac{x^2}{a^2}} = b \sqrt{1 - \frac{a^2/2}{a^2}} = b \sqrt{1/2} = \frac{b}{\sqrt{2}}\).

Step 6: Conclusion

The rectangle of greatest area that can be inscribed in the ellipse is such that its dimensions are \(x = \frac{a}{\sqrt{2}}\) and \(y = \frac{b}{\sqrt{2}}\). Therefore, the full width is \(2x = \sqrt{2}a\) and the height is \(2y = \sqrt{2}b\).

Key Concepts

Ellipse

An ellipse is a geometric shape that resembles an elongated circle. It consists of all points for which the sum of the distances to two fixed points, called the foci, is constant. This unique property allows ellipses to have different dimensions while preserving their shape. The standard equation of an ellipse centered at the origin is given by \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a\) and \(b\) are the semi-major and semi-minor axes, respectively.

• The ellipse equation helps us determine where on the x or y axes the ellipse reaches its maximum length.
• If \(a > b\), the ellipse is elongated along the x-axis. Conversely, if \(b > a\), it is elongated along the y-axis.

Understanding ellipses is essential in optimization problems, as they set the allowable boundary within which we can inscribe other shapes like rectangles.

Inscribed Rectangle

An inscribed rectangle is a rectangle placed within another shape, such that all its vertices touch the boundary of this shape. In our problem, this rectangle is placed within an ellipse, and its sides are parallel to the x and y axes.

• To maximize the rectangle's area, we must consider the limits set by the ellipse's boundary.
• Each vertex of the rectangle must satisfy the ellipse's equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).

The goal is to find the rectangle with the largest possible area that can fit inside the ellipse, touching the boundary at precisely four points.

Critical Points

Critical points are points on a graph where the derivative is zero or undefined. In calculus, finding critical points helps to determine where a function might have a maximum or minimum value. In the context of our area maximization problem, critical points are central to identifying the maximum area the rectangle can have.

• We first set up a function for the area of the rectangle in terms of one variable, say \(x\).
• The area function is \(A(x) = 4xb\sqrt{1 - \frac{x^2}{a^2}}\).
• We then compute its derivative, \(A'(x)\), to find where the slope of the tangent is zero.

By solving the equation \(A'(x) = 0\), we find the x-values that potentially give the maximum area. Determining these critical x-values is crucial in narrowing our search for the rectangle with the largest area.

Area Maximization

Area maximization involves finding the largest possible area under given constraints, such as the boundary conditions of an ellipse in this case. By deriving and analyzing the area function \(A(x)\), the process involves several steps:

• Model the problem to express the area as a function of one variable.
• Find the derivative to identify critical points where the area might be highest.
• Evaluate the critical points within the context of the constraints to determine the largest feasible area.

For the inscribed rectangle, this approach leads to discovering that both x and y dimensions of the rectangle are equal to half the semi-axes of the ellipse. Therefore, the optimal dimensions are \(x = \frac{a}{\sqrt{2}}\) and \(y = \frac{b}{\sqrt{2}}\), giving the maximum possible rectangle area within the ellipse.