Answer
The critical point occurs at $x=9$ and $\phi=\pi / 3$
Work Step by Step
We find:
\[
\begin{array}{c}
(1 / 2)(x \sin \phi)[(27-2 x)+(27-2 x+2 x \cos \phi)] =\operatorname{Area}(A)\\
27 x \sin \phi-2 x^{2} \sin \phi+x^{2} \sin \phi \cos \phi =\operatorname{Arca}(A)\\
27 x \sin \phi-2 x^{2} \sin \phi+x^{2} \sin \phi \cos \phi =f(x, \phi)\\
\sin \phi(27-4 x+2 x \cos \phi)=0=f_{x}(x, \phi) \\
x\left(27 \cos \phi-2 x \cos \phi-x \sin ^{2} \phi\right) =f_{\phi}(x, \phi)\\
x\left(27 \cos \phi-2 x \cos \phi+2 x \cos ^{2} \phi-x\right)=0=f_{\phi}(x, \phi)
\end{array}
\]
We get from eq.(1) that $\cos \phi=(4 x-27) /(2 x), x \neq 0$
From eq. $(2): \cos \phi=(4 x-27) /(2 x),$ which yields $4 x-27-x=0,$ so $x=9$ and
\[
\cos \phi=1 / 2, \phi=\pi / 3
\]
The critical point occurs at $x=9$ and $\phi=\pi / 3$
Using theorem $13.8.6,$
\[
f_{x x}\left(x_{0}, \phi_{0}\right) f_{\phi}\left(x_{0}, \phi_{0}\right)-f_{x \phi}^{2}\left(x_{0}, \phi_{0}\right)=D
\]
Here $D>0$ and $f_{x x}(9, \pi / 3)<0,$ so $f$ has a relative maximum at $(9, \pi / 3)$
