Answer to Question #182622 in Real Analysis for Raka

Question #182622

If 𝜙(x, y) = 0, show that the determinant

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fxx + λϕxx

fxy + λϕxy

ϕx

fxy + λϕxy

fyy + λϕyy

ϕy

ϕx

ϕy


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where 𝜆 is Lagrange’s multiplier, is positive, in case the function attains a maximum.


1
Expert's answer
2021-05-11T12:16:12-0400

fxx+λϕxxfxy+λϕxyϕxfxy+λϕxyfyy+λϕyyϕyϕxϕy0\begin{vmatrix} f_{xx}+\lambda \phi_{xx} & f_{xy}+\lambda \phi_{xy} & \phi_x \\ f_{xy}+\lambda \phi_{xy} & f_{yy}+\lambda \phi_{yy} & \phi_y \\ \phi_x & \phi_y & 0 \end{vmatrix}

det(fxx+λϕxxfxy+λϕxyϕxfxy+λϕxyfyy+λϕyyϕyϕxhi)=fxx+λϕxxdet(fyy+λϕyyϕyϕy0)fxy+λϕxydet(fxy+λϕyϕx0)+ϕxdet(fxy+λfyy+λϕyyϕxϕy)\det \begin{pmatrix} f_{xx}+\lambda \phi_{xx} & f_{xy}+\lambda \phi_{xy} & \phi_x\\ f_{xy}+\lambda \phi_{xy}& f_{yy}+\lambda \phi_{yy} & \phi_y \\ \phi_x&h&i\end{pmatrix}=f_{xx}+\lambda \phi_{xx} \cdot \det \begin{pmatrix} f_{yy}+\lambda \phi_{yy} &\phi_y\\ \phi_y&0\end{pmatrix}-f_{xy}+\lambda \phi_{xy}\cdot \det \begin{pmatrix}f_{xy}+\lambda&\phi_y\\ \phi_x &0\end{pmatrix}+\phi_x\cdot \det \begin{pmatrix}f_{xy}+\lambda& f_{yy}+\lambda \phi_{yy}\\ \phi_x&\phi_y\end{pmatrix}

[fxx+λϕxx(ϕy2)][fxy+λϕxy(ϕxϕy)]+[ϕx(ϕy)(fxy+λxy)(ϕy)(fyy+λyy)][f_{xx}+\lambda \phi_{xx} \cdot (-\phi_y ^2)]-[f_{xy}+\lambda \phi_{xy}\cdot (-\phi_x\phi_y) ]+[\phi_x\cdot (\phi_{y})(f_{xy}+\lambda_{xy})-(\phi_{y})(f_{yy}+\lambda_{yy})]


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