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先看一道问题的解答: 问题:x、y是实数,且满足等式3x~2+2y~2=6x,求x~2+y~2的最大值. 解由3x~2+2y~2=6x,得y~2=-3/2x~2+3x,从而K=x~2+y~2=x~2-3/2x~2+3x=1/2x~2+3x.故由-1/2<0,可知当x=3/2×(-1/2)=3时,有(x~2+y~2)_(max)=4(-1/2)×0-3~2/4(-1/2)=9/2. 这是一道在约束条件下可化为求二次函数最大值的问题.上述解题过程显然是错误的,而这种错误不易被学生所觉察,常常出现在作业中.错误的根源在于没有考虑到“约束条件”,而乱用二次函数y=ax~2+bx+c的极值公式来求在有限区间上该函数的
First look at the answer to a question: Question: x, y are real numbers, and satisfy the equation 3x~2+2y~2=6x, find the maximum value of x~2+y~2. Solution by 3x~2+2y~2 = 6x, get y~2=-3/2x~2+3x, so that K=x~2+y~2=x~2-3/2x~2+3x=1/2x~2+3x. -1/2<0, we can see that when x=3/2×(-1/2)=3, there is (x~2+y~2)_(max)=4(-1/2)×0- 3~2/4(-1/2)=9/2. This is a problem that can be reduced to the maximum of the quadratic function under constraint conditions. The above problem-solving process is obviously wrong, and this kind of error is not easy to be The students are aware that they often appear in the homework. The root cause of the error is not taking into account the “constraint,” and instead of using the quadratic function y=ax~2+bx+c the extreme value formula to find the function over a finite interval