Answer
Local minimum of $f(2,0)=\dfrac{1}{e^4}$
Work Step by Step
Given: $f_x(x,y)=(2x-4)e^{x^2+y^2-4x}=0, f_y(x,y)=2ye^{x^2+y^2-4x}=0$
Simplify the given two equations.
Critical point: $(2,0)$
In order to solve this problem we will have to apply Second derivative test that suggests the following conditions to calculate the local minimum, local maximum and saddle points of $f(x,y)$ or $f(x,y,z)$.
1. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\gt 0$ , then $f(a,b)$ is a local minimum.
2. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\lt 0$ , then $f(a,b)$ is a local maximum.
3. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \lt 0$ , then $f(a,b)$ is a not a local minimum or local maximum but a saddle point.
$D(2,0)=f_{xx}f_{yy}-f^2_{xy}=\dfrac{4}{e^8}\implies D=\dfrac{4}{e^8} \gt 0;f_{xx}=\dfrac{2}{e^4} \gt 0$
Hence, Local minimum of $f(2,0)=\dfrac{1}{e^4}$
