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# What is the minimum value of the expression 2x^2+3y^2+8x-24y+62 for real x and y?

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What is the minimum value of the expression 2x^2+3y^2+8x-24y+62 for real x and y?

Feb 26, 2019

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What is the minimum value of the expression

$$\mathbf{\huge{2x^2+3y^2+8x-24y+62}}$$

for real x and y?

$$\begin{array}{|rcll|} \hline \mathbf{f(x,y)} & \mathbf{=} & \mathbf{2x^2+3y^2+8x-24y+62} \\\\ f_x=\dfrac{\partial f} {\partial x} &=& 4x+8 \\ f_y=\dfrac{\partial f} {\partial y} &=& 6y-24 \\ f_{xx}= \dfrac{\partial^2 f} {\partial x^2} &=& 4 \\ f_{yy}= \dfrac{\partial^2 f} {\partial y^2} &=& 6 \\ f_{xy}=f_{yx}= \dfrac{\partial^2 f} {\partial x\partial y}= \dfrac{\partial^2 f} {\partial y\partial x} &=& 0 \\ \hline \end{array}$$

if $$f_{xx}f_{yy}-f^2_{xy}>0$$  either a maximum or a minimum.

Distinguish between these as follows:

if $$f_{xx}<0$$ and $$f_{yy} <0$$  then is a maximum point.

if $$f_{xx}>0$$ and $$f_{yy} >0$$  then is a minimum point.

$$\begin{array}{|rcll|} \hline && f_{xx}f_{yy}-f^2_{xy} \\ &=& 4\cdot 6 - 0^2 \\ &=& 24 \quad | \quad >0 \quad \text{either a maximum or a minimum} \\\\ && f_{xx} \\ &=& 4 \\ && f_{yy} \\ &=& 6 \quad | \quad f_{xx}>0 \text{ and } f_{yy} >0 \quad \text{it is a minimum point}\\ \hline \end{array}$$

The minimum value:

$$\begin{array}{|rcll|} \hline f_x(x,y) = 4x+8 &=& 0 \\ 4x+8 &=& 0 \\ 4x &=& -8 \\ x &=& -\dfrac{8}{4} \\ \mathbf{x_{\text{minimum}}} &\mathbf{=}& \mathbf{ -2 } \\\\ f_y(x,y) = 6y-24 &=& 0 \\ 6y-24 &=& 0 \\ 6y &=& 24 \\ y &=& \dfrac{24}{6} \\ \mathbf{y_{\text{minimum}}} &\mathbf{=}& \mathbf{ 4 } \\ \hline \end{array}$$

Feb 26, 2019