Question

A picture frame 5 ft. high hangs on a vertical wall so that the bottom edge is 4 ft. above an observer’s eye. How far (in ft.) should the observer be from the wall so that the vertical angle at the eye subtended by the picture, is greatest.

Answer 1

If the eye is x ft from the wall, then we have
θ = angle subtended by frame
Ø = angle to bottom of frame
Now consider
tanØ = 4/x
9/x = tan(θ+Ø) = (tanθ + tanØ)/(1 - tanθ tanØ)
9/x = (tanθ + 4/x)/(1 - tanθ * 4/x)
9/x - 36/x^2 tanθ = tanθ + 4/x
tanθ = 5x/(x^2+36)
sec^2θ dθ/dx = -5(x^2-36)/(x^2+36)^2
dθ/dx = 1/(tan^2θ + 1) * -5(x^2-36)/(x^2+36)^2
= -5(x^2-36)/(((5/(x^2+36))^2+1)(x^2+36)^2)
= -5(x^2-36)/(x^4+72x^2+1321)
since the denominator is never zero, dθ/dx=0 when x=6