# Find the area of the largest rectangle that can be

Harold Kessler 2022-01-07 Answered
Find the area of the largest rectangle that can be inscribed in the ellipse $$\displaystyle\frac{{x}^{{{2}}}}{{a}^{{{2}}}}+\frac{{y}^{{{2}}}}{{b}^{{{2}}}}={1}{x}$$

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## Expert Answer

temnimam2
Answered 2022-01-08 Author has 3858 answers
The given elipse is $$\displaystyle{\frac{{{x}^{{{2}}}}}{{{a}^{{{2}}}}}}+{\frac{{{y}^{{{2}}}}}{{{b}^{{{2}}}}}}={1}$$, whic means it is a 2a wide and 2b tall.
The area of the rectangle is $$\displaystyle{A}={\left({2}{x}\right)}{\left({2}{y}\right)}={4}{x}{y}$$
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
$$\displaystyle{\frac{{{y}^{{{2}}}}}{{{b}^{{{2}}}}}}={1}-{\frac{{{x}^{{{2}}}}}{{{a}^{{{2}}}}}}$$
$$\displaystyle{\frac{{{y}}}{{{b}}}}=\sqrt{{{1}-{\frac{{{x}^{{{2}}}}}{{{a}^{{{2}}}}}}}}$$
$$\displaystyle{y}={b}{\left(\sqrt{{{\frac{{{1}}}{{{a}^{{{2}}}}}}{\left({a}^{{{2}}}-{x}^{{{2}}}\right)}}}\right)}$$
$$\displaystyle={\frac{{{b}}}{{{a}}}}\sqrt{{{a}^{{{2}}}-{x}^{{{2}}}}}$$
Substitute for y in the rectunge area equation. Then find the derivative
$$\displaystyle{A}={4}{x}{y}={4}{x}{\left({\frac{{{b}}}{{{a}}}}\sqrt{{{a}^{{{2}}}-{x}^{{{2}}}}}\right)}={\frac{{{4}{b}}}{{{a}}}}{x}\sqrt{{{a}^{{{2}}}-{x}^{{{2}}}}}$$
$$\displaystyle{A}'={\frac{{{4}{b}}}{{{a}}}}{\left({\left({1}\right)}\sqrt{{{a}^{{{2}}}-{x}^{{{2}}}}}+{x}\times{\frac{{{1}}}{{{2}}}}{\left({a}^{{{2}}}-{x}^{{{2}}}\right)}^{{-{\frac{{{1}}}{{{2}}}}}}{\left(-{2}{x}\right)}\right)}$$ (product rule)
$$\displaystyle={\frac{{{4}{b}}}{{{a}}}}{\left(\sqrt{{{a}^{{{2}}}-{x}^{{{2}}}}}-{\frac{{{x}^{{{2}}}}}{{\sqrt{{{a}^{{{2}}}-{x}^{{{2}}}}}}}}\right)}$$
$$\displaystyle={\frac{{{4}{b}}}{{{a}}}}{\left({\frac{{{a}^{{{2}}}-{x}^{{{2}}}}}{{\sqrt{{{a}^{{{2}}}-{x}^{{{2}}}}}}}}-{\frac{{{x}^{{{2}}}}}{{\sqrt{{{a}^{{{2}}}-{x}^{{{2}}}}}}}}\right)}$$ (common denominator)
$$\displaystyle={\frac{{{4}{b}}}{{{a}}}}{\left({\frac{{{a}^{{{2}}}-{2}{x}^{{{2}}}}}{{\sqrt{{{a}^{{{2}}}-{x}^{{{2}}}}}}}}\right)}$$
Find where $$\displaystyle{A}'={0}$$
$$\displaystyle{0}={a}^{{{2}}}-{2}{x}^{{{2}}}$$
$$\displaystyle{2}{x}^{{{2}}}={a}^{{{2}}}$$
$$\displaystyle{x}\sqrt{{{2}}}={a}$$
$$\displaystyle{x}={\frac{{{a}}}{{\sqrt{{{2}}}}}}$$
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:
$$\displaystyle{y}={\frac{{{b}}}{{{a}}}}\sqrt{{{a}^{{{2}}}-{x}^{{{2}}}}}$$
$$\displaystyle={\frac{{{b}}}{{{a}}}}\sqrt{{{a}^{{{2}}}-{\left({\frac{{{a}}}{{\sqrt{{{2}}}}}}\right)}^{{{2}}}}}$$
$$\displaystyle={\frac{{{b}}}{{{a}}}}\sqrt{{{a}^{{{2}}}-{\frac{{{a}^{{{2}}}}}{{{2}}}}}}$$
$$\displaystyle={\frac{{{b}}}{{{a}}}}\sqrt{{{\frac{{{2}{a}^{{{2}}}}}{{{2}}}}-{\frac{{{a}^{{{2}}}}}{{{2}}}}}}$$
$$\displaystyle={\frac{{{b}}}{{{a}}}}\sqrt{{{\frac{{{a}^{{{2}}}}}{{{2}}}}}}$$
$$\displaystyle={\frac{{{b}}}{{{a}}}}\times{\frac{{{a}}}{{\sqrt{{{2}}}}}}$$
$$\displaystyle={\frac{{{b}}}{{\sqrt{{{2}}}}}}$$
Now, put x and y into the rectangle area equation
$$\displaystyle{A}={4}{x}{y}={4}\times{\frac{{{a}}}{{\sqrt{{{2}}}}}}\times{\frac{{{b}}}{{\sqrt{{{2}}}}}}={\frac{{{4}{a}{b}}}{{{2}}}}={2}{a}{b}$$

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