设x>=y>=z>=pi/12 且 x+y+z=pi/2

shingrons

求 cosx *siny*cosz     的最大值和最小值
出道题吧

dunwan2tellu

P = cos x * sin y * cos z = 1/2 * [ cos z - sin(x-y)] * cos z

如果定着 cos z 的值 ,欲得到 max P 则需有 min{sin(x-y)} = 0 ==> x = y
当 x = y ==> 2x + z = pi/2 ; P = 1/2 * (cos z)^2
那么 max P = 1/2 * (cos pi/15)^2 = 0.4665 ; x = y = 5pi/24 , z = pi/12

欲得到 min P 则须有 max{sin(x-y)} .因为 pi/2 = x+y+z >= x + pi/6 因此 x =< pi/3 , 而且 y >= pi/12 ==> x-y =< pi/3 - pi/12 = pi/4
所以 max{sin(x-y)} = sin(pi/4)
因此 min P = cos(pi/3) * sin(pi/12) * cos(pi/12) = 0.125