using symmetry, if one corner of the rectangle is at (h,k), the area is
a = 2h*2k = 4hk
Now, we know that k = 7√(1-h^2/169)
a = 4h*7√(1-h^2/169)
= 28/13 h√(169-h^2)
da/dh = 28/13 (2h^2-169)/√(169-h^2)
so, max area occurs when h = 13/√2
max area is 4 √13/2 7/√2 = √91
As expected, the corners are halfway to the ends of the ellipse.
Find the dimensions of a rectangle of maximum area with sides parallel to the coordinate axes that can be inscribed in the ellipse given by x^2/169 + y^2/49 = 1.
What would be the primary equation?
2 answers
hmmm. typo
max a = 4 13/√2 7/√2 = 91/2
max a = 4 13/√2 7/√2 = 91/2
1 answer