Question

如图,在矩形ABCD中,AB=8,BC=4,点H、F、E、G分别在AB、BC、CD、DA上,且AH=AG=CE=CF=x

Answer 1

DE=8-x,DG=4-x

y=32-（8-x）（4-x）-2x^2

=32-32+8x+4x-x^2-2x^2

=-3x^2+12x

=-3(x^2-4x)

=-3(x-2)^2+12

当x=2时面积最大,最大是12