Make a diagram, sideview, woman on left, wall on right.
let the line from the woman's eyes to the wall be x ft
draw line from there eyes to the top of the picture
let the angle of elevation of her eyes to the bottom of the picture be A
let the angle between top and bottom of picture be B
so we want angle B to be a maximum
angle B = angle (A+B - A)
tan B = tan(A+B - A)
= (tan(A+B) - tanA)/(1 + tan(A+B)tanA)
= (5/x - 2/x)/(1 + (5/x)(2/x))
= (3/x)/(x^2 + 10)/x^2)
= 3x/(x^2+10)
d(tanB)/dx = (3(x^2+10) - 3x(2x))/(x^2+10)^2
= 0 for a max of B
3x^2 + 30 - 6x^2 = 0
x^2 = 10
x = √10
Application of Derivatives and Integrals:
A sign 3 feet high is placed on a wall with its base 2 feet above the eye level of a woman attempting
to read it. How far from the wall should she stand to get the best view of the sign, that is, so that the
angle subtended at the woman’s eye by the sign is a maximum?
1 answer
2 answers