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A vertical picture is mounted on a wall so that its lower edge is 2 m…
Question
A vertical picture is mounted on a wall so that its lower edge is 2 m above the observer's eye level and its upper edge is 6 m above eye level. The observer stands a horizontal distance x metres from the wall. The angle theta subtended by the picture at the observer's eye is maximized at some distance x = d. Find d^2.
✓ Verified answer: 12checked by our engine — not a guess
Step-by-step solution
The angle is theta(x) = arctan(6/x) - arctan(2/x).
Then theta'(x) = -6/(x^2 + 36) + 2/(x^2 + 4).
Setting theta'(x) = 0: 2(x^2 + 36) = 6(x^2 + 4), so 2x^2 + 72 = 6x^2 + 24, giving 4x^2 = 48, x^2 = 12.
Since theta' changes from positive to negative here, this is the maximizing distance.
Thus d^2 = 12.
(Indeed tan(theta_max) = (6-2)/(2·sqrt(2·6)) = 4/(2·sqrt12) = 1/sqrt3, so theta_max = 30°.)
Final answer12
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