extreme f(x,y)=sqrt(2x^2+2y^2-4y+9)

解答

极值点

f (x, y) = 2 x^{2} + 2 y^{2} - 4 y + 9

极小值 (0, 1)

求解步骤

找到临界点:

(0, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{2 ( 2 x ^{2} + 7 )}{( 2 x ^{2} + 2 y ^{2} - 4 y + 9 ) 2 x ^{2} + 2 y ^{2} - 4 y + 9}

D (x, y) = \frac{28}{( 9 + 2 x ^{2} + 2 y ^{2} - 4 y ) ^{2}}

检查在每个临界点处的符号

检查临界点

(0, 1) :

极小值

极小值 (0, 1)

e x t re m e f (x) = - x^{2} - 2 x - 3

e x t re m e f (x) = x (x^{3} - x + 10)

e x t re m e f (x) = - 462 x + 64 x^{2} - 2 x^{3}

e x t re m e f (x, y) = 3 x^{3} y - 2 x^{2} y^{2} + 3 y

e x t re m e f (x) = - 5 x^{7} ln (2 x), (0, 6)