A hoarding of height 5 m is mounted on a vertical wall. The bottom...

Let the distance of the observer from the wall be $x$.

Let the angle of elevation to the top of the hoarding be $\alpha$ and to the bottom of the hoarding be $\beta$.

The height of the top of the hoarding from eye level is $4 + 5 = 9$ m, and the height of the bottom is $4$ m.

Therefore, $\tan \alpha = \dfrac{9}{x}$ and $\tan \beta = \dfrac{4}{x}$.

The angle subtended by the hoarding at the observer's eye is $\theta = \alpha - \beta$.

$\tan \theta = \tan(\alpha - \beta) = \dfrac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$

$\tan \theta = \dfrac{\dfrac{9}{x} - \dfrac{4}{x}}{1 + \left(\dfrac{9}{x}\right)\left(\dfrac{4}{x}\right)} = \dfrac{5x}{x^2 + 36}$

To maximize $\theta$, we must maximize $\tan \theta$ (since $\theta$ is acute).

$\tan \theta = \dfrac{5}{x + \dfrac{36}{x}}$

To maximize this expression, the denominator $x + \dfrac{36}{x}$ must be minimized.

Using the AM-GM inequality for positive numbers $x$ and $\dfrac{36}{x}$:

$\dfrac{x + \dfrac{36}{x}}{2} \ge \sqrt{x \cdot \dfrac{36}{x}} = 6$

$x + \dfrac{36}{x} \ge 12$

The minimum value of the denominator is achieved when $x = \dfrac{36}{x}$, which gives $x^2 = 36 \implies x = 6$.

Thus, the required distance is $6$ m.

Answer: $6$ m