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# Maximum and Minimum value問題

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

1.f(x,y)=x^2+y^2+x^2y+4

2.f(x,y)=ycosx

煩請各位大大指教!!

### 1 Answer

- Lotus_本願山彌陀講堂Lv 61 decade agoFavorite Answer
1. f(x,y)=x²+y²+x²y+4

∂f/∂x=2x+2xy=0 => x=0 or y=-1

∂f/∂y=2y+x^2=0

x=0, y=0

y=-1, x=+/-√2

∂²f/∂x²=2+2y

∂²f/∂x∂y=2x

∂²f/∂y²=2

(2階)行列式=4(1+y-x²)

點(0,0)處判別行列式=det(2, 0 //0, 2)= 4>0 , 又∂²f/∂x²=2>0, 故f(0,0)=4為極小

點(+/-√2, -1)處判別行列式<0,為saddle points

2.f(x,y)=ycos(x)

∂f/∂x=-ysin(x)=0

∂f/∂y=cos(x)=0

解聯立得 y=0, cos(x)=0

∂²f/∂x²= -ycos(x)

∂²f/∂x∂y= -sin(x)

∂²f/∂y²=0

行列式=-(sinx)²=(cosx)² -1

故critical point處行列式= -1<0,為 saddle points

saddle points (nπ+π/2, -1), n為任意整數