49. The picture is hung on the wall so that its lower edge is $a$ m, and the upper edge is $b$ m above the observer's eye. At what distance $x$ from the wall should the observer stand so that the picture is seen at the largest angle?
49. Let $A C=b, \quad A B=a, \quad A D=x$. From Fig. 30, it is clear that $b=x \operatorname{tg}(\alpha+\beta)=$ $=x \cdot \frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \operatorname{tg} \beta}$, but $\operatorname{tg} \beta=\frac{a}{x}$, then $\operatorname{tg} \alpha=$ $=\frac{b-a}{x+\frac{a b}{x}}$.

The maximum value of the viewing angle coincides with the maximum value of its tangent, i.e., with the maximum value of the fraction $\frac{b-a}{x+\frac{a b}{x}}$. Since $x \cdot \frac{a b}{x}=a b$, the minimum value of the sum $x+\frac{a b}{x}$ will be obtained when $x=\frac{a b}{x}$, i.e., $x=\sqrt{a b}$.