Question Number 4772 by Yozzii last updated on 07/Mar/16
$${Let}\:{z}={Ax}^{\mathrm{2}} +{Bxy}+{Cy}^{\mathrm{2}} .\:{Find}\:{conditions} \\ $$$${on}\:{the}\:{constants}\:{A},{B},{C}\:{that}\:{ensure} \\ $$$${that}\:{the}\:{point}\:\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\:{is}\:{a}\: \\ $$$$\left({i}\right)\:{local}\:{minimum}, \\ $$$$\left({ii}\right)\:{local}\:{maximum}, \\ $$$$\left({ii}\right)\:{saddle}\:{point}. \\ $$$$ \\ $$$$ \\ $$
Commented by 123456 last updated on 09/Mar/16
$${z}={Ax}^{\mathrm{2}} +{Bxy}+{Cy}^{\mathrm{2}} \\ $$$${z}_{{x}} =\mathrm{2}{Ax}+{By} \\ $$$${z}_{{y}} ={Bx}+\mathrm{2}{Cy} \\ $$$${z}_{{xx}} =\mathrm{2}{A} \\ $$$${z}_{{yy}} =\mathrm{2}{C} \\ $$$${z}_{{xy}} =\mathrm{0} \\ $$
Commented by 123456 last updated on 09/Mar/16
![H(x,y)= [((∂^2 f/∂x^2 ),(∂^2 f/(∂x∂y))),((∂^2 f/(∂y∂x)),(∂^2 f/∂y^2 )) ]= [((2A),0),(0,(2C)) ] (∂f/∂x)=2Ax+B (∂f/∂y)=2Cy+B (∂f/∂x)=0⇔2Ax+B=0⇔x=−(B/(2A)) (∂f/∂y)=0⇔2Cx+B=0⇔x=−(B/(2C)) A(x,y)=(∂^2 f/∂x^2 )=2A,Δ(x,y)=4AC A(x,y)>0,Δ(x,y)>0⇒A>0,C>0 A(x,y)<0,Δ(x,y)>0⇒A<0,C<0 Δ(x,y)<0⇒A<0,C>0∨A>0,C<0](https://www.tinkutara.com/question/Q4779.png)
$${H}\left({x},{y}\right)=\begin{bmatrix}{\frac{\partial^{\mathrm{2}} {f}}{\partial{x}^{\mathrm{2}} }}&{\frac{\partial^{\mathrm{2}} {f}}{\partial{x}\partial{y}}}\\{\frac{\partial^{\mathrm{2}} {f}}{\partial{y}\partial{x}}}&{\frac{\partial^{\mathrm{2}} {f}}{\partial{y}^{\mathrm{2}} }}\end{bmatrix}=\begin{bmatrix}{\mathrm{2}{A}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{2}{C}}\end{bmatrix} \\ $$$$\frac{\partial{f}}{\partial{x}}=\mathrm{2}{Ax}+{B} \\ $$$$\frac{\partial{f}}{\partial{y}}=\mathrm{2}{Cy}+{B} \\ $$$$\frac{\partial{f}}{\partial{x}}=\mathrm{0}\Leftrightarrow\mathrm{2}{Ax}+{B}=\mathrm{0}\Leftrightarrow{x}=−\frac{{B}}{\mathrm{2}{A}} \\ $$$$\frac{\partial{f}}{\partial{y}}=\mathrm{0}\Leftrightarrow\mathrm{2}{Cx}+{B}=\mathrm{0}\Leftrightarrow{x}=−\frac{{B}}{\mathrm{2}{C}} \\ $$$${A}\left({x},{y}\right)=\frac{\partial^{\mathrm{2}} {f}}{\partial{x}^{\mathrm{2}} }=\mathrm{2}{A},\Delta\left({x},{y}\right)=\mathrm{4}{AC} \\ $$$${A}\left({x},{y}\right)>\mathrm{0},\Delta\left({x},{y}\right)>\mathrm{0}\Rightarrow{A}>\mathrm{0},{C}>\mathrm{0} \\ $$$${A}\left({x},{y}\right)<\mathrm{0},\Delta\left({x},{y}\right)>\mathrm{0}\Rightarrow{A}<\mathrm{0},{C}<\mathrm{0} \\ $$$$\Delta\left({x},{y}\right)<\mathrm{0}\Rightarrow{A}<\mathrm{0},{C}>\mathrm{0}\vee{A}>\mathrm{0},{C}<\mathrm{0} \\ $$
Commented by Dnilka228 last updated on 10/Mar/16
$$\Delta\left({x}^{{y}} \right)<{a}\Rightarrow\mathrm{0}>{A},{X}<{Y}\:{if}\:{Y}=\mathrm{1} \\ $$$$\Delta\left({y}^{{x}} \right)>{a}\Rightarrow{A}<\sqrt{{a}+{b}}<{m} \\ $$$$ \\ $$
Commented by Dnilka228 last updated on 10/Mar/16
$$\alpha>\beta \\ $$$$\alpha=? \\ $$$$\beta=\alpha−\mathrm{2} \\ $$$$\alpha−\mathrm{2}=\mathrm{1} \\ $$$$\beta=\mathrm{3} \\ $$$$\alpha=? \\ $$