最大化

f=(1000+x)(1.1+0.01y)(1+0.01z)/3.5

 满足约束：

x+20y+15z-S=0

x>=0

y>=0

z>=0

y<=90

其中S是常数，该问题可以通过拉格朗日乘数法计算极值，也可以通过替换掉一个变量将问题转化成二元函数求极值

 需要注意的是求出的极值点不一定在定义域（x>=0,y>=0,z>=0,y<=90）内，由于这是一个单峰函数，当极值点不在定义域内时，f的最大值必然落在边界上，此时只需要计算边界上的最大值即可。-Maximize f = (1000+ x) (1.1+0.01 y) (1+0.01 z)/3.5 　satisfy the constraint: x+20 y+15 z-S = 0 x>　= 0 y>　= 0 z>　= 0 y <　= 90 where S is a constant, the problem can be calculated through the Lagrangian multiplier extreme, it can replace a variable is transformed into the dual function for extreme 　Note that the extreme points obtained not necessarily in the domain (x>　= 0, y>　= 0, z>　= 0, y < = 90), due this is a single peak function, when the extreme point is not within the definition of the time, f the maximum bound fall on the border, the border at this time only the maximum can be calculated.