Question

The lower edge of a 5 foot tall painting is 5 feet above your eye level. At what distance should you stand from the wall so your viewing angle of the painting is maximized?

Answer 1

x = sqrt(50) = 5sqrt(2) = 7.071 ft (to 3 decimals)

Explanation:

referring to the diagram

theta (x) = atan(10/x) - atan(5/x)

differentiate with respect to x

theta'(x) = 5/(x^2+25) - 10/(x^2+100)

For x to have an extremum (max. or min)

theta'(x) = 0 ="

5/(x^2+25) - 10/(x^2+100) = 0

transpose and cross multiply

10(x^2+25) -5(x^2+100) = 0

expand and simplify

10x^2+250 - 5x^2-500 = 0

5x^2 = 250

x^2=50

x = sqrt(50) = 5sqrt(2) = 7.071 ft (to 3 decimals)

Since we know that if x becomes large, theta will decrease, so

x = 5sqrt(2) is a maximum.