Question

A painting is hung on a wall in such a way that its upper and lower edges are 10 ft and 7 ft above the floor, respectively. How far away from the wall should an observer whose eyes are 5 ft above the floor stand in order to maximize the viewing angle? I need to explain this topic ("The Best View") in a presentation and teach it to a class.

Answer 1

I made a sketch, drew lines from her eyes to the top and bottom of the picture and called this angle a

I drew a horizontal from her eyes to the wall and called that angle of elevation b

so we have two tangent relations.

tan(a+b) = 12/x and tanb = 5/x

then a+b = arctan (12/x) and b = arctan (5/x)

then a = a+b - b

= arctan (12/x) - arctan (5/x)

recall that d(arctan x)/dx = 1/(1 + x^2)

so da/dx = 1/(1 + 144/x^2) (-12/x^2) - 1/(1 + 25/x^2) (-5/x)^2

= -12/(x^2 + 144) + 5/(x^2 + 25)

setting this equal to zero and simplifying gave me

12x^2 + 300 = 5x^2 + 720

7x^2 = 420

x^2 = 60

x = 2√15

so a+b = arctan (12/2√15) = ...

a+b = approximately (57.158°)

b = arctan (5/2√15) = ...

b = approximately (32.842°)

the "best" angle a is 24.32 , when she stands √60 or 7.74 ft from the wall.

Thumbs up if you like the explanation.