Question

一道数学难题已知四个非负实数x,y,z,u,满足3x+2y+z=6,2x+y-3u=1,则S=6u-z+1的最大值为

Answer 1

z=6-3x-2y>=0;

3μ=2x+y-1>=0;

线性域为以(2,0),(0,3),(0,1),(0.5.0)为顶点的四边形.

S=2*3μ-z+1=4x+2y-2-6+3x+2y+1=7x+4y-7;

S1=7*2+4*0-7=7;

S2=7*0+4*3-7=5;

S3=7*0+4*1-7=-3;

S4=7*0.5+4*0-7=-3.5;

Smax=S1=7.