不等式

Hsuan

没想到什么特别好的方法。$z+y<= 16 ;0<=y<=3 ;0<= z-3y<=8 ;$所以，$\frac{z-8}{3}<=y ,z+\frac{z-8}{3}<= z+y<=16 ,z<=14.$此时，$y=\frac{z-8}{3}=2.$