设x≥0,y≥0,2x y=6,则u=4x² 3x y²-6x-3y的最大值是
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由|X-2|+(Y+1/4)^2=0绝对值大于等于0平方也是大于等于0所以X=2Y=-1/4带入后面要求的式子得到等于0-17/8
设实数xy满足x+y-3≤0,y-x/2≥0,x-1≥0,则求u=y/x-x/y的取值范围x+y-3≤0.(1)y-(x/2)≥0,.(2)x-1≥0.(3)化出三条直线:x+y-3=0;y-(x/2
画出线性约束条件下的可行域,如图阴影部分,再作出直线y=2x,向下平移,过A点时,满足截距最大,而-z最大,即Z最小,此时z=2*1-1=1,c即为所求.如有不清楚
xy-12=4x+y≥2√(4xy)=4√(xy)xy-4√(xy)-12≥0(√(xy)-6)(√(xy)+2)≥0√(xy)≤-2,√(xy)≥6因为√(xy)≥0所以√(xy)≥6xy≥36所以
Cov(x,y)=EXY-EXEY挨个求出来不就可以了吗?EXY=1/3EY=3/5Ex=2/5Cov(x,y)=7/75
x=1-2y,所以8xy+4y²+1=8(1-2y)×y﹢4y²+1=﹣12y²+8y+1y在[0,0.5]对称轴是1/3所以y=⅓时,最大值为7/3y=0时
左式可化为[(xy)^3+(xz)^3+(yz)^3]/xyz+6xyz;然后[(xy)^3+(xz)^3+(yz)^3]/xyz>=3xyz(这一步是将分子利用(a+b+c)>=3*(abc)^(1
∵x>0,y>0,∴x+y≥2xy(当且仅当x=y时取等号),则xy≤x+y2,xy≤(x+y)24,∵x+y+xy=2,∴xy=-(x+y)+2≤(x+y)24,设t=x+y,则t>0,代入上式得,
2x+6y=6x=3-3yx≥03-3y≥0y≤10≤y≤1将x=3-3y代入u=4x^2+3xy+y^2-6x-3y=28y^2-48y+18=28(y-6/7)^2-18/7当y=0时,最大值为1
用y替换x,得到y的取值范围,代入得到边际值,再用z对y求导等于0,如果有多解,用二阶导正负判定,若为负则为局部最大值(localmaximum).再比较边际值和localmaximum即可.
∵|x-2a|+(y+3)²=0∴x=2a,y=-3∵A=2x²-3xy+y²-x+2y=8a²+16a+3,B=4x²-6xy+2y²+3
由|x-2a|+(y+3)*=0,x=2a,y=-3.B-2A=-x-5y=a.所以a=5,x=10.A=200+90+9-20-6=273.口算的,验一下.
|x-2a|+(y+3)^2=0x=2a,y=-3B-2A=aa=4x^2-6xy+2y^2-3x-y-2(2x^2-3xy+y^2-x+2y)=4x^2-6xy+2y^2-3x-y-4x^2+6xy
(x-1)(y-1)≥2xy-(x+y)+1≥2xy≥1+x+y1)xy≥1+√(x^2+y^2+2xy)xy-1≥√(x^2+y^2+2xy)(xy)^2-2xy+1≥x^2+y^2+2xy>=4x
因为所证式子及已知中x,y,z可以轮换,即性质等价,所以不妨设x>=y>=z>=0;由x+y+z=1得z=yz+xz+(1/3)xy>=0x=1,y=z=0时可取等,左边得证.又xy+yz+xz-2x
|x-3a|+(y+3)=0,所以X=3a,Y=-3把A=2x-3xy+y+x-3y,B=4x-6xy+2y+4x-y代入B-2A=a,得2X+5Y=a代入X=3a,Y=-3,得a=3所以x=9,y=