A tapestry 7 ft high hangs on a wall. Lower edge is 9 ft above observer's eye. How far from wall should observer stand to obtain most favorable view?
In other words, what distance from the wall maximizes the visual angle of observer?
Let the observer stand at a distance x from the wall. Also, let the line drawn from the point where the observer is standing to the lower edge make an angle x with the ground and the angle made by a similar line drawn to the upper edge be y.
So the angle of view of the observer is y-x.
Consider angle x, tan x is equal to 9/x. Similarly tan y = (9+7)/x.
Now we can write x= arctan (9/x) and y= arctan (16/x). The angle of observation, y-x is equal to arctan (16/x) - arctan (9/x). We need to maximize this.
So we differentiate [arctan (16/x) - arctan (9/x)] with respect to x.
We get [9 / (1+ 81/x^2)*x^2] - [16 / (1+ 256/x^2)*x^2].
Equating this to 0, we get [9 / (1+ 81/x^2)*x^2] - [16 / (1+ 256/x^2)*x^2] = 0
=> [9 / (1+ 81/x^2)*x^2] = [16/ (1+ 256/x^2)*x^2]
Cancel common terms and rearrange them.
=> 9*(1+ 256/x^2) = 16*(1+ 81/x^2)
=> 9+9*256/x^2 = 16 + 16*81/x^2
=> (1/x^2) [9*256 - 16*81] = 16-9
=> (1/x^2)*1008 = 7
=> (1/x^2) = 7/1008
=> x^2 = 1008/7
=> x = sqrt 144
=> x = 12
Therefore the optimum distance the observer should stand at to get the best view of the tapestry is 12 ft from the wall.