calculus
posted by rachel .
a billboard 20ft tall is located on top of a building with its lower edge 60 ft above the level of a viewers eye.
how far from a point directly below the sign should a viewer stand to maximize the angle (theta) between the lines of sight of the top and bottom of the billboard.
hint: let theta=ab then tan (theta)=tan(ab)=(tan(a)tan(b))/(1+tan(a)tan(b)) differentiate implicitly with respect to x and set d(theta)/dx=0.
im going to try to draw the drawing but no promises on how it'll turn out.
theta=t
........a_
......./20
....../ ..
...../t/b.
..../\/...
..././...60
,,/\/.....
./\/\a....
0./.\b\...
xc i hope that makes since!!(:

Using the hint,
tanθ = tan(ab) = (tan a  tan b)/(1 + tana tanb)
If the viewer is at a distance x,
tan a = 80/x
tan b = 60/x
tanθ = (80/x  60/x)/(1 + 80/x * 60/x)
tanθ = 20x/(x^2 + 4800)
sec^2θ θ' = 20(4800x^2)/(4800 + x^2)^2
θ' = 20(4800x^2)/((4800 + x^2)^2 * sec^2θ)
Now,
sec^2θ = 1 + tan^2θ = 1 + 40x^2/(4800+x^2)^2
so,
θ' = 20(4800x^2)/(4800 + x^2)^2 * (4800+x^2)^2/(40x^2 + (4800+x^2)^2)
= 20(4800x^2)/(40x^2 + (4800+x^2)^2)
θ' = 0 when 4800 = x^2 or,
x = 40√3 = 69.38'
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