haha
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haha asked in 科學數學 · 1 decade ago

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)

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    ∂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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  • 1 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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