Let $x$ satisfy the inequality $$2 \left( \log _ { \frac { 1 } { 3 } } x \right) ^ { 2 } + 9 \log _ { \frac { 1 } { 3 } } x + 9 \leqq 0 .$$ We are to find the maximum value of the function $$f ( x ) = \left( \log _ { 3 } x \right) \left( \log _ { 3 } \frac { x } { 3 } \right) \left( \log _ { 3 } \frac { x } { 9 } \right) .$$ The range of values of $x$ satisfying (1) is $$\mathbf { A } \sqrt { \mathbf { B } } \leqq x \leqq \mathbf { C D } .$$ When we set $t = \log _ { 3 } x$, the range of values of $t$ is $$\frac { \mathbf { E } } { \mathbf { F } } \leqq t \leqq \mathbf { G } .$$ When we express the right side of (2) in terms of $t$ and consider it as a function $g ( t )$, its derivative is $$g ^ { \prime } ( t ) = \mathbf { H } t ^ { 2 } - \mathbf { I } t + \mathbf { J } .$$ Hence $f ( x )$ is maximized at $x = \mathbf { K L }$, and its maximum value is $\mathbf { M }$.
Let $x$ satisfy the inequality
$$2 \left( \log _ { \frac { 1 } { 3 } } x \right) ^ { 2 } + 9 \log _ { \frac { 1 } { 3 } } x + 9 \leqq 0 .$$
We are to find the maximum value of the function
$$f ( x ) = \left( \log _ { 3 } x \right) \left( \log _ { 3 } \frac { x } { 3 } \right) \left( \log _ { 3 } \frac { x } { 9 } \right) .$$
The range of values of $x$ satisfying (1) is
$$\mathbf { A } \sqrt { \mathbf { B } } \leqq x \leqq \mathbf { C D } .$$
When we set $t = \log _ { 3 } x$, the range of values of $t$ is
$$\frac { \mathbf { E } } { \mathbf { F } } \leqq t \leqq \mathbf { G } .$$
When we express the right side of (2) in terms of $t$ and consider it as a function $g ( t )$, its derivative is
$$g ^ { \prime } ( t ) = \mathbf { H } t ^ { 2 } - \mathbf { I } t + \mathbf { J } .$$
Hence $f ( x )$ is maximized at $x = \mathbf { K L }$, and its maximum value is $\mathbf { M }$.