Answer
The critical point and its nature:
$\begin{array}{*{20}{c}}
{{\rm{Critical}}}&{}\\
{{\rm{Point}}}&{{\rm{Type}}}\\
{\left( {1,\frac{1}{2}} \right)}&{{\rm{local{\ } maximum}}}
\end{array}$
Work Step by Step
We have $f\left( {x,y} \right) = \ln x + 2\ln y - x - 4y$.
The partial derivatives are
${f_x} = \frac{1}{x} - 1$, ${\ \ \ }$ ${f_y} = \frac{2}{y} - 4$
${f_{xx}} = - \frac{1}{{{x^2}}}$, ${\ \ }$ ${f_{yy}} = - \frac{2}{{{y^2}}}$, ${\ \ }$ ${f_{xy}} = 0$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = \frac{1}{x} - 1 = 0$, ${\ \ }$ ${f_y} = \frac{2}{y} - 4 = 0$
The solutions are $x=1$ and $y = \frac{1}{2}$.
So, there is only one critical point: $\left( {1,\frac{1}{2}} \right)$.
We use the Second Derivative Test to determine the nature of the critical point. The results are listed in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}&{{\rm{Discriminant}}}&{}\\
{{\rm{Point}}}&{ - \frac{1}{{{x^2}}}}&{ - \frac{2}{{{y^2}}}}&0&{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\rm{Type}}}\\
{\left( {1,\frac{1}{2}} \right)}&{ - 1}&{ - 8}&0&8&{{\rm{local{\ } maximum}}}
\end{array}$
