A rectangle in an ellipse

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Quandaries & Queries

Question from David, a parent:

I need to find the max area of a rectangle inscribed in an ellipse with the equation
x²+4y²=4. What I have so far is f(x,y)=4xy
g(x,y)=x²+4y²-4=0,

y=sqrtx²-4/4
f'(x)=2x²/sqrt-4x²+2(sqrt-4+x²).

What I need to know is how to finish the problem and find the actual max area of the rectangle.
Thanks

David

Hi David,

What you have done so far is good but you have some slips in the algebra. You have x²+4y²-4=0 and solve for y to get

y=sqrt((4 - x²)/4) = 1/2 × sqrt(4 - x²)

I added parentheses to make sure what is under the root sign. Then you substitute this value for y into the area function to get

f(x) = 4xy = 4x × 1/2 × sqrt(4 - x²) = 2x sqrt(4 - x²)

This gives

f '(x) = 2 sqrt(4 - x²) - 2x²/sqrt(x² - 4)

To complete the problem you need to set f '(x) = 0 and solve for x. Setting f '(x) = 0 gives

2 sqrt(4 - x²) = 2x²/sqrt(4 - x²)

If you now multiply both sides by sqrt(x² - 4) you get

2 (4 - x²) = 2 x²

Solve for x.

Penny

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