The maximum inscribed rectangle problem of ellipse Make an inscribed rectangle in the ellipse of x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 to maximize the area of the rectangle My approach is: Let the vertex coordinates of the rectangle in the first quadrant be (acosx, bsinx) Then the length of the rectangle is 2acosx and the width is 2bsinx, Area = 4absinxcosx = 2absin2x Then, when x = 45 degrees, the area is the largest, then, when the area of the inscribed rectangle is the largest, don't they all become square? The length and width of the rectangle are √ 2a and √ 2B The maximum area of rectangle is s = 2p * 2q = 2Ab ", which is wrong?

The maximum inscribed rectangle problem of ellipse Make an inscribed rectangle in the ellipse of x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 to maximize the area of the rectangle My approach is: Let the vertex coordinates of the rectangle in the first quadrant be (acosx, bsinx) Then the length of the rectangle is 2acosx and the width is 2bsinx, Area = 4absinxcosx = 2absin2x Then, when x = 45 degrees, the area is the largest, then, when the area of the inscribed rectangle is the largest, don't they all become square? The length and width of the rectangle are √ 2a and √ 2B The maximum area of rectangle is s = 2p * 2q = 2Ab ", which is wrong?

When x = 45 degrees, the area is the largest. At this time, the coordinates of the vertex in the first quadrant are (√ 2A / 2, √ 2B / 2). Find out the slope of the line between the vertex and the origin = (√ 2B / 2) / (√ 2A / 2) = B / a ＜ 1. That is to say, the rectangle is not a square. Why is it x = 45 degrees