Maximizing the Area of a Rectangle Inscribed in an Ellipse

Imagine a geometry student who wants to draw the largest possible rectangle inscribed in a given ellipse. The ellipse's equation is x^2 4y^2 49. This article will guide you through the process of finding the area of the largest rectangle that can be inscribed in this ellipse, combining mathematical techniques with detailed steps and explanations.

Understanding the Problem

The given equation of the ellipse is x^2 4y^2 49. To understand the properties of this ellipse, it is first rewritten in standard form: [frac{x^2}{49} frac{y^2}{frac{49}{4}} 1]

This tells us that the semi-major axis 7 lies along the x-axis and the semi-minor axis 3.5 lies along the y-axis.

Step 1: Define the Rectangle

To find the area of the largest rectangle inscribed in the ellipse, we need to define the vertices of the rectangle. Suppose the vertices in the first quadrant are at (x, y). The area A of the rectangle can be expressed as:

A 2x cdot 2y 4xy.

Step 2: Express y in Terms of x

We can express y in terms of x using the ellipse's equation. Starting with:

frac{x^2}{49} frac{y^2}{frac{49}{4}} 1

We simplify this to:

frac{x^2}{49} frac{4y^2}{49} 1

Multiplying through by 49:

x^2 4y^2 49

We solve for y^2 to get:

4y^2 49 - x^2

Thus:

y^2 frac{49 - x^2}{4}

Taking the square root, we have:

y frac{7}{2}sqrt{1 - frac{x^2}{49}}

Step 3: Substitute y into the Area Formula

Substituting the expression for y into the area formula, we get:

A 4x cdot frac{7}{2}sqrt{1 - frac{x^2}{49}} 14xsqrt{1 - frac{x^2}{49}}

Step 4: Maximize the Area

To find the maximum area, we differentiate A with respect to x and set the derivative to zero. We rewrite the area function as:

A(x) 14xsqrt{1 - frac{x^2}{49}}

Using the product rule and chain rule, we differentiate A(x) and simplify:

A'(x) 14left(sqrt{1 - frac{x^2}{49}} - frac{x^2}{49sqrt{1 - frac{x^2}{49}}}right)

Setting the derivative equal to zero:

0 14left(sqrt{1 - frac{x^2}{49}} - frac{x^2}{49sqrt{1 - frac{x^2}{49}}}right)

This simplifies to:

14sqrt{1 - frac{x^2}{49}} frac{14x^2}{49sqrt{1 - frac{x^2}{49}}}

Cancelling common factors:

49 - x^2 x^2

Solving for x:

49 2x^2 Rightarrow x^2 frac{49}{2} Rightarrow x frac{7}{sqrt{2}}

Step 5: Find y

Substituting x frac{7}{sqrt{2}} back into the equation for y gives us:

y frac{7}{2}sqrt{1 - frac{left(frac{7}{sqrt{2}}right)^2}{49}}

y frac{7}{2}sqrt{1 - frac{49/2}{49}} frac{7}{2}sqrt{frac{1}{2}} frac{7}{2} cdot frac{1}{sqrt{2}} frac{7sqrt{2}}{4}

Step 6: Calculate the Area

Finally, we can calculate the area:

A 4xy 4 cdot frac{7}{sqrt{2}} cdot frac{7sqrt{2}}{4} 49

Conclusion: The area of the largest rectangle that can be inscribed in the ellipse x^2 4y^2 49 is 49.