Rectangle in Ellipse

Geometry Level 3

Find the area of the greatest rectangle that can be inscribed in in the ellipse x 2 a 2 + y 2 b 2 = 1 \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 .

Take a , b > 0 a,b> 0 .

For more problems try my set

a b \sqrt ab 2 a b 2 ab a b \frac{a}{b} a b ab

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2 solutions

Otto Bretscher
Apr 4, 2016

Let me recycle an idea I have used before, showing that we don't even need calculus.

We can parameterize the ellipse as x = a cos ( t ) x=a\cos(t) and y = b sin ( t ) y=b\sin(t) so that the area of an inscribed rectangle is A = 4 x y = 4 a b cos t sin t = 2 a b sin ( 2 t ) A=4xy=4ab\cos t\sin t=2ab\sin(2t) for 0 < t < π 2 0<t<\frac{\pi}{2} . The maximum of 2 a b \boxed{2ab} is attained when t = π 4 t=\frac{\pi}{4} .

Good, parametrization makes the solution even faster and more elegant - how I expected from you!

Andreas Wendler - 5 years, 2 months ago

Yes. This is the shortest way :)

Pulkit Gupta - 5 years, 2 months ago

You can see the solution in Lagrange multipliers .

In that page, there also is a problem asking for the greatest volume of a parallelipiped than can be inscribed in a elipsoid.

And you can generalize this problem, for example, to a finite elliptic cylinder in R 3 \mathbb{R}^3

You mustn't take a sledgehammer to crack a nut! Simple differential calculation is sufficient to maximize 4 x y in the ellipse.

Andreas Wendler - 5 years, 2 months ago

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If you read a little bit of this article, this example can be solved with simple differential calculation.

Guillermo Templado - 5 years, 2 months ago

Simple trigonometry nails it but ;)

Pulkit Gupta - 5 years, 2 months ago

"Mit Kanonen auf Spatzen schiessen," wie wir in der Schweiz sagen ;)

Otto Bretscher - 5 years, 2 months ago

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