Question

Area of the greatest rectangle that can be inscribed in an ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1\) is (a) \(a b\) (b) \(2 \mathrm{ab}\) (c) \((\mathrm{a} / \mathrm{b})\) (d) \(\sqrt{(a b)}\)

Short answer

The area of the greatest rectangle that can be inscribed in the given ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1\) is \(2ab\), which corresponds to option (b).

Step-by-step solution

Express the Sides of the Rectangle in terms of a and b

Let's consider a rectangle of sides 2x and 2y inscribed inside the ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1\). We know that all the points on the ellipse satisfy the expression given. Therefore, since the corners of the rectangle lie on the ellipse, the expressions x and y must satisfy the ellipse equation. Now that we have defined that, let's express the relationship between x, y, a, and b: \[\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1\]

Find the Area of the Rectangle

Since we need to find the area of the largest rectangle, we need to calculate the area of the given rectangle and maximize it. The area of the rectangle is given by: \[A = (2x)(2y) = 4xy\]

Use Calculus to Maximize the Area

To find the maximum area, we can use calculus. To do this, first express y in terms of x using the ellipse equation: \[y = b\sqrt{1 - \frac{x^2}{a^2}}\] Now plug this expression for y into our area equation: \[A = 4x\left(b\sqrt{1 - \frac{x^2}{a^2}}\right)\] Now, we will find the critical points of A(x) by calculating the derivative with respect to x and setting it equal to zero: \[\frac{dA}{dx} = 4b\left(\sqrt{1 - \frac{x^2}{a^2}}\right) - \frac{8bx^3}{a^2\sqrt{1 - \frac{x^2}{a^2}}}\] Set this equal to zero and solve for x: \[4b\left(\sqrt{1 - \frac{x^2}{a^2}}\right) - \frac{8bx^3}{a^2\sqrt{1 - \frac{x^2}{a^2}}} = 0\] Simplifying this expression, we find that \(x = \frac{a}{\sqrt{2}}\). Now we can find the corresponding y value by substituting this x value into our ellipse equation: \[y = b\sqrt{1 - \frac{\left(\frac{a}{\sqrt{2}}\right)^2}{a^2}} = b\sqrt{1 - \frac{1}{2}} = \frac{b}{\sqrt{2}}\]

Calculate the Maximum Area

Finally, we can calculate the maximum area using the x and y values we found: \[A_{max} = 4\left(\frac{a}{\sqrt{2}}\right)\left(\frac{b}{\sqrt{2}}\right) = 2ab\] Therefore, the area of the greatest rectangle that can be inscribed in the given ellipse is \(2ab\), which corresponds to option (b).

Key concepts

Ellipse Equation

An ellipse is a geometric shape that is essentially an elongated circle. The standard form of an ellipse equation is given by:

• \( \left(x^{2} / a^{2}\right) + \left(y^{2} / b^{2}\right) = 1 \)

In this equation, \(a\) and \(b\) represent the semi-major and semi-minor axes respectively. The ellipse is centered at the origin of a coordinate plane, and its shape is determined by these two axes.
The longer axis, with length \(2a\), is known as the major axis, while the shorter one, \(2b\), is the minor axis. Any point \((x, y)\) on the ellipse satisfies this equation. This relation is fundamental to solving problems that involve shapes inscribed within an ellipse, like a rectangle. Understanding this equation helps us explore more complex concepts, such as the optimization of inscribed shapes.

Calculus Optimization

Optimization in calculus revolves around finding maximum or minimum values of a function. When dealing with problems related to inscribed shapes in ellipses, optimization helps in determining the size and dimensions that maximize (or minimize) some property, often the area.
To optimize using calculus, we:

• Express the quantity to be maximized or minimized as an equation.
• Use derivatives to find critical points, which are potential maximum or minimum values.

In our example, we want to find the largest area of a rectangle that can fit inside an ellipse. By expressing the area of the rectangle and using the constraint from the ellipse equation, we employ calculus techniques to find where this area is maximized. Understanding this process provides a deeper insight into how derivatives can be used to solve real-world geometric problems.

Inscribed Shapes

Inscribed shapes are figures placed within another shape so that all vertices of the inscribed shape touch the boundary of the outer shape. Calculating the dimensions and properties of these shapes involves understanding geometric constraints and sometimes using calculus for optimization.
In our scenario, the focus is on a rectangle inscribed in an ellipse. This means that each corner of the rectangle rests precisely on the ellipse. The challenge here is not only to ensure the rectangle fits within the ellipse, but also to determine its largest possible area.
By analyzing inscribed shapes, one can explore diverse geometric and algebraic relationships. It reveals how different dimensions relate and how they affect the area and space within the encapsulating shape.

Critical Points Calculation

Critical points in calculus are found by taking the derivative of a function and setting it to zero. They represent the points where there is a potential maximum or minimum value. To confirm whether these points are indeed maxima or minima, further tests are used, such as the second derivative test.
In the exercise of finding the largest inscribed rectangle, critical points help identify where the rectangle achieves its maximum area. We start by:

• Calculating the derivative of the area function with respect to one variable.
• Setting the derivative equal to zero to find possible critical points.
• Substituting these values back to assess which yields the greatest area.

The process is a fundamental part of calculus, showcasing how derivatives are powerful tools in solving optimization problems in geometry.