Harold Kessler

2022-01-07

Find the area of the largest rectangle that can be inscribed in the ellipse $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1x$

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$A=\left(2x\right)\left(2y\right)=4xy$

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

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

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

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

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

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

$A=4xy=4x\left(\frac{b}{a}\sqrt{{a}^{2}-{x}^{2}}\right)=\frac{4b}{a}x\sqrt{{a}^{2}-{x}^{2}}$

${A}^{\prime}=\frac{4b}{a}(\left(1\right)\sqrt{{a}^{2}-{x}^{2}}+x\times \frac{1}{2}{({a}^{2}-{x}^{2})}^{-\frac{1}{2}}(-2x))$ (product rule)

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

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

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

Find where${A}^{\prime}=0$

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

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

$x\sqrt{2}=a$

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

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:

The area of the rectangle is

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

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

Find where

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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