Question

如图,在一个直角三角形的内部作矩形ABCD,其中AB和AD在两直角边上,设AB=x cm,矩形ABCD的面积为ycm2,

Answer 1

∵BC//AF,

∴△EBC∽△EAF,

BC/AF=BE/AE,

AD=BC,

AD/30=（40-x)/40,

∴AD=(3/4)(40-x)（cm).

y=3x(40-x)/4

=30-3x^2/4=-(3/4)(x^2-40x+400)+300=-(3/4)(x-20)^2+300,

∴当x=20cm时,y有最大值为300cm^2.