extreme f(p,q)=p^2-2p+0.8q^2

解答

极值点

f (p, q) = p^{2} - 2 p + 0.8 q^{2}

极小值 (1, 0)

求解步骤

找到临界点:

(1, 0)

\frac{\partial ^{2} f}{\partial p ^{2}} = 1.6

D (p, q) = 3.2

检查在每个临界点处的符号

检查临界点

(1, 0) :

极小值

极小值 (1, 0)

e x t re m e f (x) = x^{3} - 3 x^{2} + 12, - 2 \leq x \leq 4

e x t re m e f (x) = \frac{1}{2} ln (\frac{1 + x}{1 - x}) - 2 x

e x t re m e f (x) = 6 x + 12 y - x^{2} - 3 y^{2}

e x t re m e f (x) = x^{\frac{2}{5}} (x - 8)

e x t re m e f (x) = (\frac{x}{2}) + \frac{1}{x}