Find the critical points of the following functions. Use the Second Derivative Test to determine...

Step 1
Differentiate ${\displaystyle f(x,y)={(4x-1)}^2+{(2y+4)}^2+1}$ partially with respect to x and y to obtain the values of ${\displaystyle f_x}$ and ${\displaystyle f_y}$.
${\displaystyle f(x,y)={(4x-1)}^2+{(2y+4)}^2+1}$
${\displaystyle f_x=\frac{\partial {(4x-1)}^2+{(2y+4)}^2+1}{\partial x}}$
${\displaystyle =32x-8}$
${\displaystyle f_y=\frac{\partial {(4x-1)}^2+{(2y+4)}^2+1}{\partial y}}$
${\displaystyle =8(y+2)}$
Step 2
Equate ${\displaystyle f_x}$ and ${\displaystyle f_y}$ to 0 and solve for x and y.
${\displaystyle f_x=0}$
${\displaystyle 32x-8=0}$
${\displaystyle x=\frac{8}{32}}$
${\displaystyle =\frac{1}{4}}$
${\displaystyle f_y=8(y+2)}$
${\displaystyle 0=8(y+2)}$
${\displaystyle y=-2}$
The critical point is ${\displaystyle (\frac{1}{4},-2)}$.
Step 3
Differentiate ${\displaystyle f_x}$ and ${\displaystyle f_y}$ partially with the respect to x and y respectively to obtain the value of ${\displaystyle f_{xx}}$ and ${\displaystyle f_{yy}}$ and differentiate ${\displaystyle f_x}$ with respect to y to obtain the value of ${\displaystyle f_{xy}}$ equate ${\displaystyle f_{xx}}$ and ${\displaystyle f_{yy}}$ to A and C respectively and ${\displaystyle f_{xy}}$ to B.
${\displaystyle f_{xx}=\frac{\partial (32x-8)}{\partial x}}$
${\displaystyle =32}$
${\displaystyle A=32}$
${\displaystyle f_{yy}=\frac{\partial \left(8(y+2)\right)}{\partial y}}$
${\displaystyle =8}$
${\displaystyle C=8}$
${\displaystyle f_{xy}=\frac{\partial (32x-8)}{\partial y}}$
${\displaystyle =0}$
${\displaystyle B=0}$
Step 4
Substitute the values of A,B,C in ${\displaystyle AC-B^2}$ and obtain the value of the expression.
${\displaystyle AC-B^2=\left(32\right)\left(8\right)-0^2}$

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