Let-z-Ax-2-Bxy-Cy-2-Find-conditions-on-the-constants-A-B-C-that-ensure-that-the-point-0-0-0-is-a-i-local-minimum-ii-local-maximum-ii-saddle-point-

Question Number 4772 by Yozzii last updated on 07/Mar/16

L e t z = {A x}^{2} + B x y + {C y}^{2} . F i n d c o n d i t i o n s

o n t h e c o n s t a n t s A, B, C t h a t e n s u r e

t h a t t h e p o i n t (0, 0, 0) i s a

(i) l o c a l m i n i m u m,

(i i) l o c a l m a x i m u m,

(i i) s a d d l e p o i n t .

Commented by 123456 last updated on 09/Mar/16

z = {A x}^{2} + B x y + {C y}^{2}

z_{x} = 2 A x + B y

z_{y} = B x + 2 C y

z_{x x} = 2 A

z_{y y} = 2 C

z_{x y} = 0

Commented by 123456 last updated on 09/Mar/16

H (x, y) = [\begin{matrix} \frac{\partial^{2} f}{\partial x^{2}} & \frac{\partial^{2} f}{\partial x \partial y} \\ \frac{\partial^{2} f}{\partial y \partial x} & \frac{\partial^{2} f}{\partial y^{2}} \end{matrix}] = [\begin{matrix} 2 A & 0 \\ 0 & 2 C \end{matrix}]

\frac{\partial f}{\partial x} = 2 A x + B

\frac{\partial f}{\partial y} = 2 C y + B

\frac{\partial f}{\partial x} = 0 \Leftrightarrow 2 A x + B = 0 \Leftrightarrow x = - \frac{B}{2 A}

\frac{\partial f}{\partial y} = 0 \Leftrightarrow 2 C x + B = 0 \Leftrightarrow x = - \frac{B}{2 C}

A (x, y) = \frac{\partial^{2} f}{\partial x^{2}} = 2 A, Δ (x, y) = 4 A C

A (x, y) > 0, Δ (x, y) > 0 \Rightarrow A > 0, C > 0

A (x, y) < 0, Δ (x, y) > 0 \Rightarrow A < 0, C < 0

Δ (x, y) < 0 \Rightarrow A < 0, C > 0 \lor A > 0, C < 0

Commented by Dnilka228 last updated on 10/Mar/16

Δ (x^{y}) < a \Rightarrow 0 > A, X < Y i f Y = 1

Δ (y^{x}) > a \Rightarrow A < \sqrt{a + b} < m

Commented by Dnilka228 last updated on 10/Mar/16

α > β

α = ?

β = α - 2

α - 2 = 1

β = 3

α = ?