如图,将长方形OABC放在指教直角坐标系中,o为原点,点A在y轴正半轴上,点E是AB上的一动点

3个回答

  • (1)设A(0,a),B(b,a),C(b,0)

    反比例函数y=k/x与AB(y = a)的交点为E(k/a,a)

    反比例函数y=k/x与BC(x = b)的交点为F(b,k/b)

    AE = k/a,CF = k/b

    S1 S2 = (1/2)*OA*AE (1/2)OC*CF

    = (1/2)a*k/a (1/2)b*k/b)

    = k/2 k/2

    = k = 2

    k = 2

    (2)设A(0,2),B(4,2),C(4,0)

    E(k/2,2),F(4,k/4)

    AE = k/2,EB = 4 - k/2

    CF = k/4,FB = 2 - k/4

    四边形OAEF的面积=

    矩形OABC的面积 - 三角形OCF的面积 - 三角形BEF的面积

    = 4*2 - (1/2)OC*CF - (1/2)EB*FB

    = 8 - (1/2)*4*k/4 - (1/2)(4 - k/2)(2 - k/4)

    = 8 - k/2 -(1/2)(8 - k - k k²/8)

    = 4 k/2 - k²/16

    = 5 - (k - 4)²/16

    k = 4时,四边形OAEF的面积最大,为5.

    此时E(2,2)