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17世纪英国数学家瓦里斯提出一个问题:周长相等的所有矩形中,以正方形的面积最大,证明这个问题的方法很多,这里我们采用二次函数的最大值的方法来解.设矩形的周长为2a,一边长为x,则另一边长为a-x,故矩形面积为:S=x(a-x)=-x~2+ax=-(x-a/2)~2+a~2/4.可见,当x=a/2时,S_(最大)=a~2/4.此时,另一边长a-x=a-a/2=a/2.该矩形是正方形.这就是说:“周长一定的矩形中,以正方形的面积最大.”利用二次函数把瓦里斯问题层层引申,可得到许多有趣的数学问题.
The seventeenth-century British mathematician Variss asked a question: in all the rectangles with equal circumference, the square area is the largest. There are many ways to prove this problem. Here we use the maximum of the quadratic function to solve. Length of 2a, while the length of x, the other side of the length of ax, so the rectangular area is: S = x (ax) = - x ~ 2 + ax = - (xa / 2) ~ 2 + a ~ 2 / Visible, when x = a / 2, S_ (maximum) = a ~ 2 / 4. At this time, the other side of the long ax = aa / 2 = a / 2. The rectangle is a square. This means: In a certain rectangle, the square has the largest area. "The use of quadratic functions to extend the Wallis problem leads to many interesting mathematical problems.