What is the minimum value of the expression $2x^2+3y^2+8x-24y+62$ for real $x$ and $y$?
+6248 \(2x^2 + 3y^2 + 8x - 24y + 62 = \\ 2(x^2 +4x) + 3(y^2 -8y) + 62 = \\ 2(x^2+4x+4-4) + 3(y^2 -8y+16-16) + 62 =\\ 2(x+2)^2-8 + 3(y-4)^2 -48 + 62 =\\ 2(x+2)^2 +3(y-4)^2 +6 \\ \text{which is clearly minimum at }x=-2,~y=4 \\ \text{and has the minimum value of }6\)
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+26367 What is the minimum value of the expression 2x^2+3y^2+8x-24y+62 for real x and y?
\(\begin{array}{|rclrcl|} \hline f(x,y) &=& 2x^2+3y^2+8x-24y+62 \\ f_x &=& 4x+8 \quad & \quad f_y &=& 6y-24 \\ f_x &=& 0 \quad & \quad f_y &=& 0 \\ 4x+8 &=& 0 \quad & \quad 6y-24 &=& 0 \\ x &=& \dfrac{-8}{4} \quad & \quad y &=& \dfrac{24}{6} \\ x &=& -2 \quad & \quad y &=& 4 \\ && \text{The critical Point is: } (-2,4). \\ \hline \end{array}\)
\(\begin{array}{|l|} \hline \text{Find the nature of the critical Point, we apply the second derivative test: } \\ \quad \text{If } f_{xx}\cdot f_{yy} - \left(f_{xy}\right)^2 > 0 \text{ and } f_{xx} > 0 \text{ minimum point } \\ \quad\text{If } f_{xx}\cdot f_{yy} - \left(f_{xy}\right)^2 > 0 \text{ and } f_{xx} < 0 \text{ maximum point } \\ \quad\text{If } f_{xx}\cdot f_{yy} - \left(f_{xy}\right)^2 < 0 \text{ saddle point } \\ \quad f_{xx} = 4, ~ f_{xy}=0, ~ f_{yy} = 6 \\ \quad f_{xx}\cdot f_{yy} - \left(f_{xy}\right)^2 = 4\cdot 6-0^2 = 24 > 0 \text{ and } f_{xx} = 4 > 0 \text{ minimum point } \\ \hline \end{array}\)
+26367 How is \(\mathbf{f_{xy}}\) calculated ?
\(\boxed{f(x,y) = 2x^2+3y^2+8x-24y+62 } \\ \small{ \begin{array}{|rcl|rcl|} \hline \dfrac{\partial f}{\partial x} = &f_x& = 4x + 8 \quad & \quad \dfrac{\partial f}{\partial y} = &f_y& = 6y-24 \\ \dfrac{ d\left(\dfrac{\partial f}{\partial x}\right) }{\partial x}= \dfrac{\partial^2 f}{\partial x\partial x} = &f_{xx}& = 4 ~ & ~ \dfrac{ d\left(\dfrac{\partial f}{\partial y}\right) }{\partial y}= \dfrac{\partial^2 f}{\partial y\partial y} = &f_{yy}& = 6 \\\\ \dfrac{ d\left(\dfrac{\partial f}{\partial x}\right) }{\partial y}= \dfrac{\partial^2 f}{\partial x\partial y} = &f_{xy}& = \dfrac{ d(4x+8)}{\partial y} = 0 ~ & ~ \dfrac{ d\left(\dfrac{\partial f}{\partial y}\right) }{\partial x}= \dfrac{\partial^2 f}{\partial y\partial x} = &f_{yx}& = \dfrac{d(6y-24)}{\partial x} = 0 \\\\ \hline \end{array}}\\ \boxed{f_{xy} = f_{yx}} \\ \)
