Y=26100-XZ=(1400+X*3)(1.4+(X/100+Y/20)/100)问:当X取值多少时,Z有最大值我的

1个回答

  • z = (1400 + x³)[1.4 + (x/100 + y/20)/100]

    = (1400 + x³)[1.4 + x/10000 + y/2000]

    = (1400 + x³)[1.4 + x/10000 + (26100 - x)/200]

    = (1400 + x³)[14.45 - x/2500]

    = 20230 - 0.56x + 14.45x³ - x⁴/2500

    dz/dx = -0.56 + 43.35x² - x³/625

    d²z/dx² = 86.7x - 3x²/625

    令 dz/dx = 0,

    解得:x₁= 27093.75,x₂= 0.11366,x₃= -0.11366

    当 x₁= 27093.75,d²z/dx² = -1174514.063 < 0

    当 x₂= 0.11366,d²z/dx² = 9.8543 〉 0

    当 x₃= -0.11366,d²z/dx² = -0.9854 < 0

    当 x = 27093.75,

    z = [20230 - 0.56x + 14.45x³ - x⁴/2500]|(x=27093.75) = 7.1848×10¹³

    当 x = -0.11366

    z = [20230 - 0.56x + 14.45x³ - x⁴/2500]|(x=-0.11366) = 20230.042

    所以,当 x = 27093.75,z(max) = 7.1848×10¹³