Question

A painting is hung on a wall in such a way that its upper and lower edges are 10 ft and 7ft above the floor, respectively. An observer whose eyes are 5 ft above the floor stands x feet from the wall. How far away should the observer stand to maximize the angle subtended by the painting

Answer 1

You paint over the picture to mae it more sufficient and what not to prepare to do water colors like the old man did u know hat i mean hahaha but ya I'm really good at those kind of questions ?

Answer 2

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

(check my arithmetic, especially for the derivative da/dx
I should have written this out on paper first)

so a+b = arctan (12/2√15) = ...
a+b = appr57.158°
b = arctan 5/2√15 = ...
b = appr 32.842

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