设Q:x²+y²≤1,0≤z≤1,则Jf(x+y+z)dV=().

设变量x,y满足约束条件{x-y≥0,x+y≤1,x+2y≥0},则函数Z=2x+y的最大值?设变量x,y满足约束条件{x-y≥0,x+y≤1,x+2y≥0},则函数Z=2x+y的最大值?设变量x,y

设变量x,y满足约束条件{x-y≥0,x+y≤1,x+2y≥0},则函数Z=2x+y的最大值?设变量x,y满足约束条件{x-y≥0,x+y≤1,x+2y≥0},则函数Z=2x+y的最大值?设变量x,y

设x,y满足约束条件y≥0,x≥y,x+y≤1,则z=2x+y的最大值是设x,y满足约束条件y≥0,x≥y,x+y≤1,则z=2x+y的最大值是设x,y满足约束条件y≥0,x≥y,x+y≤1,则z=2

设x,y满足约束条件(x+y-7≤0),(x-3y+1≤0),(3x-y-5≥0),则z=2x-y的最大值为?设x,y满足约束条件(x+y-7≤0),(x-3y+1≤0),(3x-y-5≥0),则z=

设xy满足约束条件x≤0x≤y2x-y+1≥0则z=3x+2y的最大值设xy满足约束条件x≤0x≤y2x-y+1≥0则z=3x+2y的最大值设xy满足约束条件x≤0x≤y2x-y+1≥0则z=3x+2

设x,y满足约束条件:x-y+1≥0,x+y-1≥0,x≤3.则z=2x-3y的最小值为?设x,y满足约束条件:x-y+1≥0,x+y-1≥0,x≤3.则z=2x-3y的最小值为?设x,y满足约束条件

设x,y满足约束条件x≥0x-2y≥0x-y≤1,则z=2x+y的最大值设x,y满足约束条件x≥0x-2y≥0x-y≤1,则z=2x+y的最大值设x,y满足约束条件x≥0x-2y≥0x-y≤1,则z=

设变量x,y,z满足x+y+z=10≤x≤10≤y≤23y+z≥0求F=3x+6y+4z的最大值和最小值设变量x,y,z满足x+y+z=10≤x≤10≤y≤23y+z≥0求F=3x+6y+4z的最大值

设X,Y满足约束条件X－Y＋1≥0X＋Y－1≥0X≤3,则Z=2X－3Y的最小值是设X,Y满足约束条件X－Y＋1≥0X＋Y－1≥0X≤3,则Z=2X－3Y的最小值是?（线性问题）设X,Y满足约束条件X

设x,y,z是正实数,且x+y+z=1.求证：(1)xy+yz+xz≤1/3,(2)x√y+y√z+z√x≤√3/3.2小时内采纳.过时无用.要用柯西不等式的方法做的设x,y,z是正实数,且x+y+z

设实数x,y满足2x+y≤4,x-y≥-1,x-2y≤2,则z=x+y的最大值为设实数x,y满足2x+y≤4,x-y≥-1,x-2y≤2,则z=x+y的最大值为设实数x,y满足2x+y≤4,x-y≥-

设x、y满足{2x+y≥4,x-y≥-1,x-2y≤2,则z=x+y设x、y满足{2x+y≥4,x-y≥-1,x-2y≤2,则z=x+y设x、y满足{2x+y≥4,x-y≥-1,x-2y≤2,则z=x

设x,y,z满足约束条件x+y+z=1,3y+z≥2,0≤x≤1,0≤y≤1,求u=2x+6y+4z的最大值和最小值.6,4)设x,y,z满足约束条件x+y+z=1,3y+z≥2,0≤x≤1,0≤y≤

设变量x,y,z满足约束条件：x+y+z=1,0≤x≤1,0≤y≤2,3y+z≥2,求F=3x+6y+4z的最大值.设变量x,y,z满足约束条件：x+y+z=1,0≤x≤1,0≤y≤2,3y+z≥2,

设x,y满足约束条件x+y≤1,y≤x,y≥-2,则z=3x+y的最大值为设x,y满足约束条件x+y≤1,y≤x,y≥-2,则z=3x+y的最大值为设x,y满足约束条件x+y≤1,y≤x,y≥-2,则

设x,y满足约束条件x+y≤1,y≤x,y≥-2,则z=3x+y的最大值设x,y满足约束条件x+y≤1,y≤x,y≥-2,则z=3x+y的最大值设x,y满足约束条件x+y≤1,y≤x,y≥-2,则z=

设x>=0,y>=0,z>=0,p=-3x+y+2z,q=x-2y+4z,x+y+z=1,求点(p,q)活动范围过程,谢谢设x>=0,y>=0,z>=0,p=-3x+y+2z,q=x-2y+4z,x+

设变量x,y满足约束条件2x+y≤2,x≥y,y≥-1,则z=-y+3x的最大值为?设变量x,y满足约束条件2x+y≤2,x≥y,y≥-1,则z=-y+3x的最大值为?设变量x,y满足约束条件2x+y

设x,y满足约束条件{x-y+5≥0,x+y≥0,x≤3}求z=(x+1)^2-y^2的最大值?设x,y满足约束条件{x-y+5≥0,x+y≥0,x≤3}求z=(x+1)^2-y^2的最大值?设x,y

集合与不等式问题设集合P={x∣lnx＞0},Q=｛x∣∣x+1∣≤6,x∈Z｝,则P∩Q的子集的个数为只需要给我答案!集合与不等式问题设集合P={x∣lnx＞0},Q=｛x∣∣x+1∣≤6,x∈Z｝