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# Calculus:找極大值&極小值&saddle point

Find the local maximum and minimum values and saddle points of the function.

F(x,y)=ycos(x)

### 2 Answers

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- Lotus_本願山彌陀講堂Lv 61 decade agoFavorite Answer
∂F/∂x= -ysin(x) =0

∂F/∂y= cos(x)=0

解聯立得 x=(2n-1)π/2, n為整數, y=0

又點( (2n-1)π/2, 0)處

∂^2 F/∂x^2= -ycos(x)=0

∂^2 F/∂x∂y= -sin(x)=-1 or 1

∂^2 F/∂y^2=0

判別行列式=

| 0 - 1|

| -10 |

or

| 0 1 |

| 1 0 |

行列式值均為-1

故((2n-1)x/2, 0)處為F(x,y)=y cos(x)的 saddle points

(沒有極大值,極小值)

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- 無名小子Lv 71 decade ago
F(x,y)=ycosx

Fx(x,y)=-ysinx

Fy(x,y)=cosx

Fx(x,y)=0得到-ysinx=0

Fy(x,y)=0得到cosx=0

求-ysinx=0 且 cosx=0

解為x=(2n-1)*(π/2)且y=0

Fxx(x,y)=-ycosx

Fyy(x,y)=0

Fxy(x,y)=-sinx

將x=(2n-1)*(π/2)且y=0代入

D=Fxx(x,y)*Fyy(x,y)-[Fxy(x,y)]^2

=0*0-[+/-1]^2=-1<0

因為D<0

所以F(x,y)在 ((2n-1)*(π/2),0)有saddle points

****最佳解請給Lotus兄

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