A painting in an art gallery has height h and is hung so its lower edge is a distance d above the eye of an observer. How far from the wall should the observer stand to get the best view? (In other words, where should the observer stand so as to maximize the angle \theta subtended at his eye by the painting?) (answer may depend on h and d.)

Answer 1

Draw a diagram, looking from the side. If the observer is at distance x, then

The angle Ø from the eye to the bottom of the picture is such that

tanØ = d/x

The angle θ subtending the picture is

tan(Ø+θ) = (d+h)/x

We want to determine x for maximum θ.

So, eliminating x, we get

d/tanØ = (d+h)/tan(Ø+θ)

Nah, that's getting messy. How about this: The distance a from the eye to the bottom of the picture is

a^2 = x^2+d^2

The distance b from the eye to the top of the picture is

b^2 = x^2 + (d+h)^2

Now, we want to find maximum θ using the law of cosines:

h^2 = b^2 + a^2 - 2abcosθ
h^2 = x^2+d^2 + x^2 + (d+h)^2 - 2√((x^2+d^2)(x^2+(d+h)^2)cosθ

cosθ = (h^2-(x^2+d^2)-(x^2+(d+h)^2))/(2√((x^2+d^2)(x^2+(d+h)^2))
= -2(x^2+dh+d^2)/√(x^4 + (d^2+(d+h)^2)x^2 + d^2(d+h)^2)

Messy, but if we let c = d+h, then

cosθ = -2(x^2+dc)/√(x^4 + (d^2+c^2)x^2 + d^2c^2)

-sinθ dθ/dx = 2x(c-d)^2(cd-x^2)/(junk)^3/2

The junk is that radical to the 3/2. We don't really have to worry about it, because it is never zero, and sinθ is not zero, so we just need

2x(c-d)^2(cd-x^2) = 0
2xh^2((d+h)d-x^2) = 0
2h^2x^3 - 2dh^3x - 2d^2h^2x = 0
2x(h^2x^2 - dh(h^2+dh)) = 0

x^2 = dh^2(h+d)/h^2

whew - still messy

A view of the solution using angles is at

http://www.nfva.info/LarsonCalc/Ch5_files/ch5_5_6w_files/image018.jpg

but it has actual numbers. You will have to plug in h and d for a general solution.

Answer 2

I tried method 2 but gave up about an hour ago. I do not think there is any trick, just plug and chug.