Question

求椭圆x^2/a ^2+y^/b^2=1(a>b>o)得内接矩形面积的最大值.

Answer 1

设第一象限内的交点是(x,y),(x>0,y>0),则其它点是(x,-y),(-x,-y),(-x,y)所以S=4xy1=x^2/a ^2+y^/b^2≥2√[(xy)/(ab)]^2=2xy/(ab)所以xy≤ab/2即S=4xy≤2ab此时x^2/a ^2=y^/b^2=1/2,x=(√2)a/2,y=(√2)b/2...