1,已知abc是三个非负数,并而且满足3a+2b+c=5,2a+b-3c=1设M=3a+b-7c,求M的最大值和最小值?2、实数XYZ满足X+Y+Z=5.XY+YZ+ZS=3,则Z的最大值是?

Question

1,已知a b c是三个非负数,并而且满足3a+2b+c=5,2a+b-3c=1设M=3a+b-7c,求M的最大值和最小值?
2、实数X Y Z 满足X+Y+Z=5.XY+YZ+ZS=3,则Z的最大值是?

Answer 1

1.把c当作已知数,解得
a=7c-3
b=-11c+7
∵a≥0,b≥0,c≥0
∴7c-3≥0,-11c+7≥0,c≥0
解得3/7≤c≤7/11
m=3a+b-7c=3(7c-3)+(-11c+7)-7c=3c-2
把3/7≤c≤7/11代入得-5/7≤m≤-1/11
所以m的最大值为-1/11,最小值为-5/7
2.∵xy+yz+zx=3,x+y+z=5 ∴x+y=5-z ∴2xy=6-2(y+x)z=6-2(5-z)z=2z^2-10z+6 ∴2*(xy+yz+zx)=6 ∵x+y+z=5 ∴（x+y+z）^2=25 x^2+y^2+z^2+2*(xy+xz+yz)=25 x^2+y^2+z^2=19 ∵（x-y）^2≥0,x^2+y^2-2xy≥0,x^2+y^2≥2xy,∴x^2+y^2=2xy时,z^2有最大值,∴z^2+2xy=19,∴z^2+2z^2-10z+6=19,3z^2-10z-13=0 z^2-10z/3-13/3=0 （z-5/3）^2-(8/3)^2=0 (z-13/3)*(z+1)=0 z1=13/3 z2=-1 z1>z2 故z的最大值=13/3