extreme h(v,t)=3vt-9t^2

解答

极值点

h (v, t) = 3 v t - 9 t^{2}

鞍点 (0, 0)

求解步骤

找到临界点:

(0, 0)

\frac{\partial ^{2} f}{\partial v ^{2}} = - 18

D (v, t) = - 9

检查在每个临界点处的符号

检查临界点

(0, 0) :

鞍点

鞍点 (0, 0)

e x t re m e f (x) = \frac{( 2 x - 9 )}{x}

e x t re m e f (x, y) = 400 - 9 x^{2} - 36 y^{2}

e x t re m e f (x) = (2 - x)^{2} (0 + x)

e x t re m e f (x, y) = \frac{- 4 x}{x ^{2} + y ^{2} + 1}

e x t re m e f (x, y) = x^{2} - 6 x + 10