Expert Answer to "Find the area of the largest rectangle that..."

Harold Kessler

Answered question

2022-01-07

Find the area of the largest rectangle that can be inscribed in the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 x

Answer & Explanation

temnimam2

Beginner2022-01-08Added 36 answers

The given elipse is

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

, whic means it is a 2a wide and 2b tall.
The area of the rectangle is

A = (2 x) (2 y) = 4 x y

We have to solve the ellipse equation for y, then substitute for y in the area.
We can use just the positive square root, that represents the upper half of the ellipse

\frac{y^{2}}{b^{2}} = 1 - \frac{x^{2}}{a^{2}}

\frac{y}{b} = \sqrt{1 - \frac{x^{2}}{a^{2}}}

y = b (\sqrt{\frac{1}{a^{2}} (a^{2} - x^{2})})

= \frac{b}{a} \sqrt{a^{2} - x^{2}}

Substitute for y in the rectunge area equation. Then find the derivative

A = 4 x y = 4 x (\frac{b}{a} \sqrt{a^{2} - x^{2}}) = \frac{4 b}{a} x \sqrt{a^{2} - x^{2}}

A^{'} = \frac{4 b}{a} ((1) \sqrt{a^{2} - x^{2}} + x \times \frac{1}{2} {(a^{2} - x^{2})}^{- \frac{1}{2}} (- 2 x))

(product rule)

= \frac{4 b}{a} (\sqrt{a^{2} - x^{2}} - \frac{x^{2}}{\sqrt{a^{2} - x^{2}}})

= \frac{4 b}{a} (\frac{a^{2} - x^{2}}{\sqrt{a^{2} - x^{2}}} - \frac{x^{2}}{\sqrt{a^{2} - x^{2}}})

(common denominator)

= \frac{4 b}{a} (\frac{a^{2} - 2 x^{2}}{\sqrt{a^{2} - x^{2}}})

Find where

A^{'} = 0

0 = a^{2} - 2 x^{2}

2 x^{2} = a^{2}

x \sqrt{2} = a

x = \frac{a}{\sqrt{2}}

That's a local maximum of A, and since the minimum is 0 that it's likely this is the only maximum.
Find the corresponding y value with the ellipse equation:

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