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.


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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

$α > β$ $α = ?$ $β = α - 2$ $α - 2 = 1$ $β = 3$ $α = ?$
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