kenct113 asked in 科學數學 · 1 decade ago

# 偏導函數：極大極小值

１.

f ( x , y ) = 4x^2 + xy + y^2 + 2x - y + 13，試決定有無局部極大極小？

２.

f ( x , y ) = x^2 + 4xy + y^2 + 2x - y + 13，試決定有無局部極大極小？

３.

f ( x , y ) = x^2 - y^2 + 2x - 4y + 5，試決定有無局部極大極小？

..1332

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• linch
Lv 7

1. f(x, y) = 4x^2 + xy + y^2 + 2x - y + 13

f_x(x, y) = 8x + y + 2

f_y(x, y) = x + 2y - 1

f_x = 0, f_y = 0 ==> (x, y) = (-1/3, 2/3) is a critical pont(臨界點) of f.

f_xx(x, y) = 8, f_xy(x, y) = 1, f_yy (x, y) = 2

D(x, y) = f_xx(x, y) f_yy(x, y) - f_xy(x, y)^2

D(- 1/3, 2/3) = 16 - 1 = 15 > 0 and f_xx(- 1/3, 2/3) = 1 > 0

f( - 1/3, 2/3) =

f has a local minimum value 37/3 at the point (- 1/3, 2/3).

2. f(x, y) = x^2 + 4xy + y^2 + 2x - y + 13

f_x(x, y) = 2x + 4y + 2

f_y(x, y) = 4x + 2y - 1

f_x = 0, f_y = 0 ==> (x, y) = (2/3, - 5/6) is a critical pont of f.

f_xx(x, y) = 2, f_xy(x, y) = 4, f_yy (x, y) = 2

D(x, y) = f_xx(x, y) f_yy(x, y) - f_xy(x, y)^2

D( 2/3, -5/6) = 4 - 16 = - 12 < 0 and f( 2/3, -5/6) = 169/12

( 2/3, - 5/6, 169/12) is a saddle point(鞍點). (鞍點沒有極值)

3. f(x, y) = x^2 - y^2 + 2x - 4y + 5

f_x(x, y) = 2x + 2

f_y(x, y) = - 2y - 4

f_x = 0, f_y = 0 ==> (x, y) = ( - 1, - 2) is a critical pont of f.

f_xx(x, y) = 2, f_xy(x, y) = 0, f_yy (x, y) = - 2

D(x, y) = f_xx(x, y) f_yy(x, y) - f_xy(x, y)^2

D(- 1, - 2) = - 4 < 0 and f( - 1, - 2) = 8

( - 1, - 2, 8) is a saddle point.

• 1 decade ago

3個都對f(x,y)做一次微分和二次微分

1.Df(x,y)=(8x+y+2,x+2y-1)

微分等於0===>局部極值

解聯立方程式8x+y+2=0和x+2y-1=0

解得x=-1/3,y=2/3

D²f(x,y)=

[8 1]

[1 2]

因為剛好是常數，帶入x=-1/3,y=2/3還是此2*2矩陣

行列式值det(D²f(x,y))=15大於0且特徵數eigenvalues=5+根號10和5-根號10(皆大於0)===>局部極小...Ans

如果行列式值小於0且eigenvalues皆小於0，則為局部極大