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一、数形结合优化解题例1已知2x+y=4.求x~2+y~2的最小值.常规解法由2x+y=4,得y=4-2x,所以x~2+y~2=x~2+(4-2x)~2=16-16x+5x~2=5(x-8/5)~2+16/5.于是可知当x=8/5时,x~2+y~2的最小值为16/5.反思常规解法将求代数式x~2+y~2的最小值转化为求二次函数的最小值,其中配方有一定难度.考生如果注意到(x~2+y~2)~(1/2)=((x-0)~2+(y-0)~2)~(1/2),由此联想到两点间的距离公式,就可以考虑将代数式x~2+y~2的最小值转化为距离的最小值的平方.优化解法令d=(x-0)~2+(y-0)~2,则d表示
First, the number of combinations with the optimal solution Problem 1 Known 2x + y = 4. Find the minimum x ~ 2 + y ~ 2. The conventional solution by 2x + y = 4, y = 4-2x, so x ~ 2 + y ~ 2 = x ~ 2 + (4-2x) ~ 2 = 16-16x + 5x ~ 2 = 5 (x-8/5) ~ 2 + 16 / 5. It can be seen that when x = 8/5, the minimum of x ~ 2 + y ~ 2 is 16/5. Reflecting the conventional solution will find the minimum of algebraic x ~ 2 + y ~ 2 is converted to the minimum quadratic function, in which the formula has a certain degree of difficulty. To (x ~ 2 + y ~ 2) ~ (1/2) = ((x-0) ~ 2 + (y-0) ~ 2) ~ (1/2) Formula, we can consider the minimum value of the algebraic x ~ 2 + y ~ 2 is converted to the square of the minimum distance optimization solution made d = (x-0) ~ 2 + (y-0) ~ 2, then d