Answer
$(1,4)$ is a local maximum point.
Work Step by Step
Let us first find the critical points of the function where $f_x$ and $f_y$ are equal to 0, assuming first and second partial derivatives are continuous.
$f_x=\frac{y}{2\sqrt{x}}-2$ and $f_y=\sqrt{x}-2y+7$
We can substitute $y=4\sqrt{x}$ into equation 2.
$f_y=-7\sqrt{x}+7$, which is equal to 0 when $x=1$. Thus, we have one critical point at $(1,4)$.
$f_{xx}=\frac{-y}{4x^{3/2}}$ and $f_{yy}=-2$ and $f_{xy}=\frac{1}{2\sqrt{x}}$.
$D(1,4)=(-1)(-2)-(\frac{1}{2})^2>0$ and $f_{xx}=\frac{-y}{4x^{3/2}}<0$, which means $(1,4)$ is a local maximum point.
$f(1,4)=14$
