**Problem: A 4 by 4 picture hangs on a wall such that its bottom edge is 2 ft above your eye level. ****How far back from the picture should you stand, directly in front of the picture, in order to view the picture under the maximum angle?**

The point of view is the point at eye level. The picture hangs 2ft above that point. The angle α needs to be maximized. We need to find the value d such that this is so, where d is the distant between the person and the picture. Initially, using trigonometric functions, we find that tan(β) = 2/d and tan(α + β) = (2+4)/d = 6/d.

Thus we get β = tan^{-1}(2/d) and α + β = tan^{-1}(6/d). Now solving for α, we get that α = tan^{-1}(6/d) - tan^{-1}(2/d).

Now we want to take the first derivative of α and set the derivative equal to zero. Thus, we get α' = (-6/(d^{2}+36)) - (-2/(d^{2}+4)) = (-6/(d^{2}+36)) + (2/(d^{2}+4)).

Now we set the derivative equal to zero getting, 0 = (-6/(d^{2}+36)) + (2/(d^{2}+4)). Simplifying, we get 6/(d^{2}+36) = 2/(d^{2}+4).

Continuing to simplify, we get 6(d^{2}+4) = 2(d^{2}+36), and 6d^{2}+24 = 2d^{2}+72.

Thus 4d^{2} = 48, d^{2} = 12, and d = ±2√3.

Because we are dealing with distance, we will take d = 2√3.

Now that we know the value for d, we will find the value of α which will maximize the view of the picture.

We know that α = tan^{-1}(6/d) - tan^{-1}(2/d) = Π/3 - Π/6 = Π/6 = 30°.

Therefore, the view of the picture is maximized when there is an angle measure of 30°.