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?
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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.
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