Answer
The distance $x$ that maximizes the angle $\theta$ is $\sqrt{2}~meters$
Work Step by Step
$a$ is the distance from eye level to the bottom of the picture. Thus, $a=1$
$b$ is the distance from eye level to the top of the picture. Thus, $b=2$
The optimal value of $x$ is $\sqrt{ab}$:
$\sqrt{ab} = \sqrt{(1)(2)} = \sqrt{2}$
The distance $x$ that maximizes the angle $\theta$ is $\sqrt{2}~meters$
