论文部分内容阅读
习题:已知x~2/a~2+y~2/b~2=1(a>0,b>0,x≥0,y≥0),设P=x+y,求P的最大值和最小值。此题散见于各种数学资料中,由数形结合法不难求得,P_(max)=(a~2+b~2)~(1/2),P_(min)=min(a,b),利用这一结论直接求解形如y=(ax+b)~(1/2)+(cx+d)~(1/2)(a、c<0)的函数最值将非常简捷。例1 求函数y=(5+x)~(1/2)+(4-x)~(1/2)的最大值和最小值。解:设v=(5+x)~(1/2),v=(4-x)~(1/2),则v~2=20+4x。v~2=4-x,消去x得v~2/36+v~2/9=1。∴y_(max)=45~(1/2)=5~(1/3),y_(min)=3。例2 求函数y=(ax-b)~(1/2)+(c-dx)~(1/2)(a>0,d>0,且ac>bd)的最大值和最小值。
Exercises: It is known that x~2/a~2+y~2/b~2=1 (a>0, b>0, x≥0, y≥0). Let P=x+y, find the maximum of P. Value and minimum value. This question is scattered in all kinds of mathematical data and is easily found by the combination of number and shape. P_(max)=(a~2+b~2)~(1/2), P_(min)=min(a, b) Using this result to directly solve the shape of the function such as y = (ax + b) ~ (1/2) + (cx + d) ~ (1/2) (a, c <0) will be very simple . Example 1 Find the maximum and minimum values of the function y=(5+x)~(1/2)+(4-x)~(1/2). Solution: Let v=(5+x)~(1/2), v=(4-x)~(1/2), then v~2=20+4x. v~2=4-x, eliminating x gets v~2/36+v~2/9=1. ∴y_(max)=45~(1/2)=5~(1/3), y_(min)=3. Example 2 Find the maximum and minimum values of the function y=(ax-b)~(1/2)+(c-dx)~(1/2)(a>0, d>0, and ac>bd).