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在“希望杯”试题中,有一类多元函数的条件最值问题,题目中含有xy和x2+y2两个因子,若利用-(x2+y2)/2≤xy≤(x2+y2)/2这一简单不等式来求解。简捷明快,事半功倍.
In the “Hope Cup” question, there is a conditional value problem for multivariate functions. The title contains xy and x2+y2 two factors, if -(x2+y2)/2≤xy≤(x2+y2)/2 is used This simple inequality solves. Simple and clear, more effective.