Let $a > 1$. We divide the region defined by the two inequalities $$0 \leqq x \leqq \frac { \pi } { 6 } , \quad 0 \leqq y \leqq a \cos 3 x$$ into two sections by the straight line $y = 1$. Let us denote the area of the section where $y \geq 1$ by $S$ and the area of the section where $y \leq 1$ by $T$. We are to find the value of $a$ such that $T - S$ is maximized, and also find the maximum value of $T - S$. Let $t$ denote the value of $x \left( 0 \leqq x \leqq \frac { \pi } { 6 } \right)$ satisfying the equation $a \cos 3 x = 1$. Then we have $$\begin{aligned}
S & = \frac { \sin 3 t } { \mathbf { A } \cos 3 t } - t \\
S + T & = \frac { 1 } { \mathbf { B } \cos 3 t } .
\end{aligned}$$ When we set $f ( t ) = T - S$, we see that $$f ^ { \prime } ( t ) = \frac { ( \mathbf { C } - \mathbf { D } \sin 3 t ) \sin 3 t } { \cos ^ { \mathbf { E } } 3 t } .$$ Hence $T - S$ is maximized at $t = \frac { \pi } { \mathbf { F G } }$. Thus, $T - S$ is maximized at $a = \frac { \mathbf { H } \sqrt { \mathbf { I } } } { \mathbf{J} }$, and the maximum value is $\frac { \pi } { \mathbf { K } }$.
Let $a > 1$. We divide the region defined by the two inequalities
$$0 \leqq x \leqq \frac { \pi } { 6 } , \quad 0 \leqq y \leqq a \cos 3 x$$
into two sections by the straight line $y = 1$. Let us denote the area of the section where $y \geq 1$ by $S$ and the area of the section where $y \leq 1$ by $T$. We are to find the value of $a$ such that $T - S$ is maximized, and also find the maximum value of $T - S$.
Let $t$ denote the value of $x \left( 0 \leqq x \leqq \frac { \pi } { 6 } \right)$ satisfying the equation $a \cos 3 x = 1$. Then we have
$$\begin{aligned}
S & = \frac { \sin 3 t } { \mathbf { A } \cos 3 t } - t \\
S + T & = \frac { 1 } { \mathbf { B } \cos 3 t } .
\end{aligned}$$
When we set $f ( t ) = T - S$, we see that
$$f ^ { \prime } ( t ) = \frac { ( \mathbf { C } - \mathbf { D } \sin 3 t ) \sin 3 t } { \cos ^ { \mathbf { E } } 3 t } .$$
Hence $T - S$ is maximized at $t = \frac { \pi } { \mathbf { F G } }$. Thus, $T - S$ is maximized at $a = \frac { \mathbf { H } \sqrt { \mathbf { I } } } { \mathbf{J} }$, and the maximum value is $\frac { \pi } { \mathbf { K } }$.