Rectangle inscribed in ellipse

1. Apr 7, 2010

shanshan

1. The problem statement, all variables and given/known data
Find the dimensions of the largest rectangle with sides parallel to the axes that can be inscribed in the ellipse x^2 + 4y^2 = 4

2. Relevant equations

3. The attempt at a solution
I simplified the equation of the ellipse into the ellipse formula:
x^2/4 + y^2 = 1
Then I manipulated the equation to isolate y:
y = (sqrt(4-x^2)/2)
Then, since the area of the rectangle can be divided into four equal parts with equal length and width, I substituted my y into my area formula, a = 4xy:
a = 2x(sqrt(4-x^2))
Now I find the derivative of my area...
a' = (8-4x^2)/sqrt(4-x^2)
... Set it equal to 0 to find my maximum value for x:
(4(2-x^2))/sqrt(4-x^2) = 0
And find that x is equal to sqrt(2). (which is the right answer, now I just need y)
Then, I substituted my value for x back into my area formula (a = 4x(sqrt(4-x^2)/2)))
4(sqrt(2))(sqrt(4-(sqrt(2))^2)/2)
and end up with my area as 4.
Then I substituted both this area and my x value back into my area formula, a = 4xy:
4 = 4(sqrt(2))y
and find that y is equal to 4. However, the back of my book tells me that I should have y = 2sqrt(2).
I have looked it over several times and cannot find my mistake. Help would be much appreciated!!

Last edited: Apr 8, 2010
2. Apr 7, 2010

Staff: Mentor

Make life easier on yourself, and just maximize the area of the the portion of the rectangle in the first quadrant. Due to the symmetry of the rectangle and the ellipse, the values of x and y that maximize that rectangle will also maximize the area of the larger rectangle, whose area is 4 times that of the smaller rectangle.

3. Apr 8, 2010

shanshan

Okay, I will keep that in mind for next time...
But either way, I come out with the same answer, y = 4.
I'm thinking maybe the answer key is wrong, I've done this question so many times!

4. Apr 8, 2010

Staff: Mentor

Mistake above. 4 = 4*sqrt(2)*y ==> y = 1/sqrt(2) = sqrt(2)/2

So if x = sqrt(2) and y = sqrt(2)/2, then A = 4xy = 4*sqrt(2)*sqrt(2)/2 = 4
If the book has this, then it's wrong. The dimensions of the rectangle are 2x = 2sqrt(2) and 2y = sqrt(2). The ellipse has its major axis along the x-axis and its minor axis along the y-axis. The vertices of the ellipse are at (+/-2, 0) and (0, +/-1).

5. Apr 8, 2010

shanshan

Okay. I had sqrt(2)/2 on my sheet as my answer, I guess I just typed it wrong.
But Thankyou!