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/(d2+36)) - (-2/(d2+4)) = (-6/(d2+36)) + (2/(d2+4)).

Now we set the derivative equal to zero getting, 0 = (-6/(d2+36)) + (2/(d2+4)). Simplifying, we get 6/(d2+36) = 2/(d2+4).

Continuing to simplify, we get 6(d2+4) = 2(d2+36), and 6d2+24 = 2d2+72.

Thus 4d2 = 48, d2 = 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°.


Return to main