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非线形规划问题

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realnumber Posted at 2013-8-26 20:05:46 |Read mode
\begin{cases} x\ge y \\ x\le 3 \\ x+y\le 4 \\ y\ge 0   \end{cases}
则$Z=x^2+5y$的最大值是什么?
----可以解吗?如果线形约束条件个数再多几个如何?

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kuing Posted at 2013-8-26 20:14:01
y 是一次的,只要考虑上面的边界,然后变成分段函数求最大值……

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 Author| realnumber Posted at 2013-8-26 20:17:49
虽然不是很明白,那修改,如果这样$Z=-(x-1)^2-3(y-1)^2$在(1,1)最大了,即使这个点不在边界,
现在改为$Z=x^2+2xy+2y^2+5y$数据是胡乱写的,想到办法的网友可以自行修改数据或形式,

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kuing Posted at 2013-8-26 20:46:13
两个元都不是一次的当然就不能直接判断是边界上了,次数越高越复杂

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hongxian Posted at 2013-8-26 23:06:11
回复 3# realnumber


    $y$是一次的为什么在边界取最值,大概是因为将$x$看着是固定的,则$z$是关于$y$的一次函数的缘故。

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 Author| realnumber Posted at 2013-8-27 07:50:50
回复 5# hongxian
恩,一下子没反应过来,后来想通了

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