设X≥Y≥Z≥π/12,且X+Y+Z=π/2求乘积cosXsinYcosZ的最大值和最小值

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设X≥Y≥Z≥π/12,且X+Y+Z=π/2求乘积cosXsinYcosZ的最大值和最小值
设X≥Y≥Z≥π/12,且X+Y+Z=π/2
求乘积cosXsinYcosZ的最大值和最小值

设X≥Y≥Z≥π/12,且X+Y+Z=π/2求乘积cosXsinYcosZ的最大值和最小值
首先,Z不会小于π/12,不会大于π/6,其次,
cosX递减,而sinY递增,那么对于任何的Z,当cosXsinYcosZ取得极大值时,必定cosX结果为正,并且cosX越大时,sinY越大,所以必然X=Y,
cosXsinYcosZ化为
cosXsinXcos(π/2-2X)
=-1/2sin2Xcos2X
=-1/4sin4X
X取值范围为[π/6,5π/24],4X取值为[2π/3,5π/6],该范围包含3π/4
所以cosXsinYcosZ最大值为1/4,此时X=Y=3π/16π,Z=π/8
cosXsinYcosZ取得最小时,由于X可取得[π/6,π/3],
所以cosX必为负,且此条件下:
X越大,结果越小
Y越大,结果越小
Z越大,结果越大,
所以Z应该取尽量小,这个取值的影响只会让结果越小,而不会变大,所以X+Y=5π/12
cosXsinYcosZ化为
cos(X)sin(Y)*√3/2
=√3/4*(sin(5π/12)+sin(2Y-5π/12))(积化和差公式)
Y取值为[π/12,π/6],2Y-5π/12为[-π/4,-π/12],当Y等于π/12时,sin(2Y-5π/12)取得极小值-1
cosXsinYcosZ极小值等于
√3/4*(1/2-1)=-√3/8
此时Z=Y=π/12,X=π/3

太难了

用拉格朗日乘法
u=cosXSINYCOSZ-M(X+Y+Z-π/2)
F(X))=-SINYCOSZSINX-M
F(Y)=COSXCOSYCOSZ-M
F(Z)=-SINYCOSXSINZ-M
F(M)=X+Y+Z-π/2
X=Z=π/12 Z=π/3

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设X≥Y≥Z≥π/12，且X+Y+Z=π/2
15度<=z<=y<=x
求乘积cosXsinYcosZ的最大值和最小值
cosXsinYcosZ
=(1/2)sin(y+z)[2sinycosz]
＝(1/2)sin(y+z)[sin(y+z)+sin(y-z)]
>=(1/2)[sin(y+z)]^2
>=(1/2)sin(π/6)sin(π/6)
=1/8
cosXsinYcosZ
=(1/2)sin(y+z)[2sinycosz]
＝(1/2)sin(y+z)[sin(y+z)+sin(y-z)]
=(1/2)cosx[cosx+sin(y-z)]
对于给定的x
sin(y-z)max时
π/3>=x>=5π/24时
(1/2)cosx[cosx+sin(y-z)]<=(1/2)cosx[cosx+sin(π/3-x)]
π/6y=x
z=π/2-2x
y-z=3x-π/2
(1/2)cosx[cosx+sin(y-z)]<=(1/2)cosx[cosx+sin(3x-π/2)]
＝(1/2)cosx[cosx－cos3x]
两个一元函数，自己作吧！！

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最大值1最小值0

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