Question Locate and describe the relative extrema, if any exist, of the function f(x, y) = 3x² + 3y2 – 12x + 12y + 5 The critical point produces a Select an answer Hint: Enter an ordered triplet (2, y, z) into the final box. Find the critical point of the function f(x, y) = 1+ 3x – 7x2 + 6y + 4y2 = This critical point is a: Minimum

\begin{array}{l}f(x, y)=3 x^{2}+3 y^{2}-12 x+12 y+5 \\f x=\frac{\partial}{\partial x}\left[3 x^{2}+3 y^{2}-12 x+12 y+5\right] \\f x=6 x-12 \\f y=\frac{\partial}{\partial y}\left[3 x^{2}+3 y^{2}-12 x+12 y+5\right] \\f y=6 y+12 \\-x=0 \\; \quad f y=0 \\6 x-12=0 \\6 y+12=0 \\x=2 \\y=-2 \\(x, y)=(2,-2) \\f_{x x}=\frac{\partial}{\partial x}[6 x+12]=6=r \\f_{y y}=\frac{\partial}{\partial y}(6 y+12)=6=t \\f x y=\frac{\partial}{\partial y} \cdot(6 x-12)=0=S \\r t-s^{2}=6(6)-0=36 \\\gamma t-s^{2}>0 ; \quad \gamma>0 \\\end{array}\therefore(2,-2) is local minimaat x=2, y=-2\begin{aligned}f(x, y) &amp; =3(u)+3(u)-24-24+5 \\&amp; =24-24-24+5 \\&amp; =-19\end{aligned}\therefore(2,-2) is critical point(2,-2,-19) is local minimacSScanned with CamScanner\begin{array}{l}f(x, y)=1+3 x-7 x^{2}+6 y+4 y^{2} \\f x=\frac{\partial}{\partial x}\left(1+3 x-7 x^{2}+6 y+4 y^{2}\right) \\f x=3-14 x \\f y=\frac{\partial}{\partial y}\left(1+3 x-7 x^{2}+6 y+4 y^{2}\right) \\f y=6+5 y \\f_{x}=0 \quad ; f_{y}=0 \\3-14 x=0 \\6+8 y=0 \\y=-3 / 4 \\(x, y)=\left(\frac{3}{14}, \frac{-3}{4}\right) \text { is critical point } \\\end{array}\begin{array}{l} f_{x x}=\frac{\partial}{\partial x}(3-14 x)=-14=r \\f_{y y}=\frac{\partial}{\partial y}(6+8 y)=8=t \\f_{x y}=\frac{\partial}{\partial y}(3-14 x)=0=s \\r t-s^{2}=-14(8)-0 \\=-112\end{array}r t-s^{2}<0CS Scanned with CamScanner ...