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

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
$\frac{{y}^{2}}{{b}^{2}}=1-\frac{{x}^{2}}{{a}^{2}}$
$\frac{y}{b}=\sqrt{1-\frac{{x}^{2}}{{a}^{2}}}$
$y=b\left(\sqrt{\frac{1}{{a}^{2}}\left({a}^{2}-{x}^{2}\right)}\right)$
$=\frac{b}{a}\sqrt{{a}^{2}-{x}^{2}}$
Substitute for y in the rectunge area equation. Then find the derivative
$A=4xy=4x\left(\frac{b}{a}\sqrt{{a}^{2}-{x}^{2}}\right)=\frac{4b}{a}x\sqrt{{a}^{2}-{x}^{2}}$
${A}^{\prime }=\frac{4b}{a}\left(\left(1\right)\sqrt{{a}^{2}-{x}^{2}}+x×\frac{1}{2}{\left({a}^{2}-{x}^{2}\right)}^{-\frac{1}{2}}\left(-2x\right)\right)$ (product rule)
$=\frac{4b}{a}\left(\sqrt{{a}^{2}-{x}^{2}}-\frac{{x}^{2}}{\sqrt{{a}^{2}-{x}^{2}}}\right)$
$=\frac{4b}{a}\left(\frac{{a}^{2}-{x}^{2}}{\sqrt{{a}^{2}-{x}^{2}}}-\frac{{x}^{2}}{\sqrt{{a}^{2}-{x}^{2}}}\right)$ (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:

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