We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
118
1
avatar

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

 Feb 25, 2019

Best Answer 

 #1
avatar+5782 
+1

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

 

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

.
 Feb 25, 2019
 #1
avatar+5782 
+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

9 Online Users