# Explain how to find the dimensions of the rectangle of maximum area that can be inscribed inside the pictured ellipse `(x/6)^2+(y/5)^2=1`

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### 1 Answer

Given the ellipse `(x/6)^2+(y/5)^2=1` , find the maximum area for an inscribed rectangle.

The area of the rectangle will be A=(2x)(2y)=4xy. (2x and 2y by symmetry -- allow one vertex to be at (x,y) )

`(y/5)^2=1-(x/6)^2`

`y/5=sqrt(1-(x/6)^2)`

`y=5sqrt(1-(x/6)^2)`

So the area is `A=4x(5sqrt(1-(x/6)^2))=20xsqrt(1-(x/6)^2)`

To maximize the area take the first derivative and set equal to zero:

`A'=20sqrt(1-(x/6)^2)+20x(1/2)(1-(x/6)^2)^(-1/2)(-x/18)`

` ` `=20sqrt(1-(x/6)^2)-5/9x^2(1-(x/6)^2)^(-1/2)`

Setting `A'=0` we get:

`20sqrt(1-(x/6)^2)=(5/9 x^2)/sqrt(1-(x/6)^2)`

`20(1-(x/6)^2)=5/9x^2`

`20-(5x^2)/9=5/9x^2`

`180-5x^2=5x^2`

`10x^2=180 ==> x=3sqrt(2),y=5/2sqrt(2)`

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**The dimensions of the rectangle are `6sqrt(2)"x"5sqrt(2)` **

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