My sketch looked like this:
Side view,
A vertical wall, T is the top and V is the bottom of the TV, Q is the point along a horizontal line level with her eyes, which I called P
(we want that distance PQ)
so we have TV = 30, and VQ = 6
Let PQ = x
let angle TPV = A ---. that's going to be a maximum angle
let angle VPQ = B
then tan B = 6/x and tan (A+B) = 36/x
tan(A+B) = ( tanA + tanB)/( 1 - tanAtanB) --->one of our identities in trig
36/x = (tanA + 6/x)/(1 - (6/x)tanA )
36 - (216/x^2)tanA = xtanA + 6/x^2
times x^2
36x^2 - 216tanA = x^3tanA + 6
x^3tanA + 216tanA = 36x^2 - 6
tanA(x^3 + 216) = 36x^2 - 6
tanA = (36x^2 - 6)/(x^3 + 216)
take derivative of both sides with respect to x
sec^2 A dA/dx = [ (x^3 + 216)(72x) - 3x^2(36x^2 - 6) ]/(x^3+216)^2
for a max of angle A, dA/dx = 0
so (x^3 + 216)(72x) - 3x^2(36x^2 - 6) = 0
3x[ 24(x^3+216) - x(36x^2 - 6) ] = 0
x = 0 , which makes no sense OR
24x^3 + 5184 - 36x^3 + 6x = 0
-12x^3 + 6x + 5184 = 0
-----2x^3 - x - 864 = 0 ----
tough to solve, so I went to Wolfram
and got x = 7.58157
http://www.wolframalpha.com/input/?i=2x%5E3+-+x+-+864+%3D+0
seems rather unreasonable, so I tested the answer by taking a value of x slightly higher and slightly lower than 7.58157
let x = 7.58157 , supposedly the best x value
tan B = 6/7.58157 , A = 38.3578°
tan (A+B) = 36/7.58157 , A+B = 78.1073
angle A = (A+B) - B =39.7495°
let x = 7.5 , slightly closer
tan B = 6/7.5 , B = 38.6598
tan(A+B) = 36/7.5 , A+B = 78.2317
angle A = 39.5719 ---- > which is a smaller angle
let x = 7.6 , slightly farther away from the TV
tanB = +/7.6, B = 38.2902°
tan(A+B) = 36/7.6 , A+B = 78.0793
angle A = 39.789 , ----> again, slightly smaller angle
Looks like my answer of 7.58157 inches gives us the largest viewing angle.
WOW, nice question.
Television
Jim’s mother watches a lot of TV. The screen is 30 inches tall and the bottom of the screen is 6 inches above her eye level while sitting in her favorite chair. To optimize her viewing experience, Jim wishes to calculate how far from the screen she should sit. Find the horizontal distance from the TV’s wall to her eyes that maximizes the angle her eyes trace from the bottom to the top of the screen.
1.draw pictures of at least THREE different cases (including the endpoints if any).
2. define all variables CLEARLY (with words) and then use those variables consistently.
3. Employ calculus to find the OPTIMAL case (which will be either a critical point or an endpoint).
1 answer