Question

area of the greatest rectangle that can be inscribed in an ellipse xsquare by a square plus y square by b square is what?

Answer 1

See the image.

Area  = (2a cos x )(2b sin x)

dA / dx = 4ab cos 2x 

For this to be max or min, dA/dx  = 0

cos 2x = 0
2x = π/2
x = π/4

Now, lets see d²A/dx² = -8ab sin 2x
When x = π/4, d²A/dx² <0
So, area is max when x = π/4

Area = 2ab sin 2x = 2ab sin π/2 = 2ab


Max area is 2ab

Answer 2

2ab is the area of the greatest rectangle that can be inscribed in an ellipse x²/a² + y²/b² = 1

Step-by-step explanation:

complete question :  ellipse x²/a² + y²/b² = 1

reffer attached diagram

Area of rectangle = 4xy

A = 4xy

dA/dx = 4xdy/dx  + 4y

x²/a² + y²/b² = 1

=> 2x/a²  + 2y(dy/dx)/b² = 0

=> dy/dx   = - xb²/a²y

dA/dx = 4x( - xb²/a²y)  + 4y

put dA/dx = 0

=> a²y² = b²x²

=> x²/a² = y²/b²

x²/a² + y²/b² = 1

=> x²/a² = y²/b²  = 1/2

x = a/√2  , y = b/√2

Area = 4xy  =  4 ( a/√2)(b/√2)  = 2ab

2ab is the area of the greatest rectangle that can be inscribed in an ellipse x²/a² + y²/b² = 1

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