Question

Area of greatest rectangle that can be inscribed in the ellipse

Answer 1

Step-by-step explanation:

Suppose that the upper righthand corner of the rectangle is at the point ⟨x,y⟩. Then you know that the area of the rectangle is, as you say, 4xy, and you know that

x2a2+y2b2=1.(1)

Thinking of the area as a function of x, we have

dAdx=4xdydx+4y.

Differentiating (1) with respect to x, we have

2xa2+2yb2dydx=0,

so

dydx=−b2xa2y,

and

dAdx=4y−4b2x2a2y.

Setting this to 0 and simplifying, we have y2=b2x2a2. From (1) we know that

y2=b2−b2x2a2.

Thus, y2=b2−y2, 2y2=b2, and y2b2=12. Clearly, then, x2a2=12 as well, and the area is maximized when

x=a2–√=a2–√2andy=b2–√=b2–√2.

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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

refer 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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