A painting of height 3 feet hangs on the wall of a museum, with the bottom of the  painting 6 feet above the floor. If the eyes of an observer are 5 feet above the floor, then,  the observer should stand x feet from the base of the wall to maximize his angle of vision  $\theta$ , where x is ….

Let x  be the distance from the observer to the base of the wall, and let  ${\theta }_0$  be the angle  between the line of sight of the bottom of the painting and the horizontal. Then  $\tan (\theta +{\theta }_0)=4/x$   and  $\tan {\theta }_0=1/x$ . Hence,  $\tan \theta =\frac{\tan (\theta +{\theta }_0)-\tan {\theta }_0}{1+\tan (\theta +{\theta }_0)\tan {\theta }_0}=\frac{4/x-1/x}{1+4/x^2}=\frac{3x}{x^2+4}$  . Since maximizing  $\theta$  is equivalent  to maximizing  $\tan \theta$, it suffices to do the latter. Now,  $D_x\left(\tan \theta \right)=3\left[\frac{(x^2+4)-x\left(2x\right)}{{(x^2+4)}^2}\right]=\frac{3(4-x^2)}{{(x^2+4)}^2}$  . Hence, the unique positive critical  number is  $x=2$ . The first-derivative test shows this to be a relative maximum, and, by  the uniqueness of the positive critical number, this is an absolute maximum.