设xy满足约束条件x+y≥1 x-y≥-1 2x-y≤2若目标函数z+4ax+3by(a>0 b>0)最大值为12则1/a+1/b最小值为

问题描述:

设xy满足约束条件x+y≥1 x-y≥-1 2x-y≤2若目标函数z+4ax+3by(a>0 b>0)最大值为12则1/a+1/b最小值为

解做出x+y≥1 x-y≥-1 2x-y≤2的可行域
知当x=3,y=4时,z=4ax+3by有最大值12
即12a+12b=12
即a+b=1
即(1/a+1/b)
=(1/a+1/b)*1
=(1/a+1/b)*(a+b)
=1+1+b/a+b/a
≥2+2√b/a*b/a
=4
J即1/a+1/b最小值为4