   SEARCH HOME Math Central 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     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.