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Find the minimum value of \(x^2 + 2xy + 3y^2 - 6x - 2y\)over all real numbers \(x\) and \(y.\)

Guest Feb 25, 2019

#1**+1 **

\(f=x^2+2xy+3y^2-6x-2y\\ \nabla f=(2x+2y-6,2x+6y-2)\\ \nabla f = 0 \Rightarrow\\ 2x+2y=6\\ 2x+6y=2\\ 4y=-4\\ y=-1,~x=4\)

\(\text{This is a stationary point. We don't know it's a minimum.}\\ \text{We have to apply the 2nd derivative test in 2D}\\ f_{xx}=2\\ f_{yy}= 6\\ f_{xy}=2\\ \left|\begin{pmatrix}2 &2\\2&6\end{pmatrix}\right|= 8 > 0 \text{ and } f_{xx}=2>0\\ \text{Thus we have a minimum}\)

.Rom Feb 25, 2019

#1**+1 **

Best Answer

\(f=x^2+2xy+3y^2-6x-2y\\ \nabla f=(2x+2y-6,2x+6y-2)\\ \nabla f = 0 \Rightarrow\\ 2x+2y=6\\ 2x+6y=2\\ 4y=-4\\ y=-1,~x=4\)

\(\text{This is a stationary point. We don't know it's a minimum.}\\ \text{We have to apply the 2nd derivative test in 2D}\\ f_{xx}=2\\ f_{yy}= 6\\ f_{xy}=2\\ \left|\begin{pmatrix}2 &2\\2&6\end{pmatrix}\right|= 8 > 0 \text{ and } f_{xx}=2>0\\ \text{Thus we have a minimum}\)

Rom Feb 25, 2019