# calculus

posted by on .

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=a-b then tan (theta)=tan(a-b)=(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\.|..|
|-----x-|c i hope that makes since!!(:

• calculus - ,

Using the hint,

tanθ = tan(a-b) = (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(4800-x^2)/(4800 + x^2)^2
θ' = 20(4800-x^2)/((4800 + x^2)^2 * sec^2θ)
Now,
sec^2θ = 1 + tan^2θ = 1 + 40x^2/(4800+x^2)^2
so,
θ' = 20(4800-x^2)/(4800 + x^2)^2 * (4800+x^2)^2/(40x^2 + (4800+x^2)^2)
= 20(4800-x^2)/(40x^2 + (4800+x^2)^2)

θ' = 0 when 4800 = x^2 or,
x = 40√3 = 69.38'