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).

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.

To solve the problem, let's follow the steps given:

1. Draw pictures of at least THREE different cases (including the endpoints if any):

Case 1: When the horizontal distance from the TV's wall to her eyes is minimal.
In this case, let's assume the horizontal distance from the TV's wall to her eyes is zero, i.e., she is sitting right in front of the TV. The angle her eyes trace from the bottom to the top of the screen will be at its maximum.

Case 2: When the horizontal distance from the TV's wall to her eyes is moderate.
In this case, let's assume the horizontal distance from the TV's wall to her eyes is halfway between the TV and her favorite chair. This distance will vary based on the proportions of the room and the placement of the TV.

Case 3: When the horizontal distance from the TV's wall to her eyes is maximum.
In this case, let's assume the TV is placed against the wall, and she is sitting in her favorite chair at a distance where the angle her eyes trace from the bottom to the top of the screen is at its minimum.

2. Define all variables CLEARLY (with words) and then use those variables consistently:

Let:
- h be the height of the TV screen (30 inches),
- d be the horizontal distance from the TV's wall to her eyes,
- e be the height of Jim's mother's eyes above the ground (eye level),
- c be the distance between the bottom of the TV screen and her eyes (6 inches).

3. Employ calculus to find the OPTIMAL case (which will be either a critical point or an endpoint):

To 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, we need to find the derivative of the angle with respect to the horizontal distance and then solve for the critical points (where the derivative is zero) or check the endpoints.

Let's denote the angle her eyes trace from the bottom to the top of the screen as θ.

From the picture, we can see that the angle θ is given by:
θ = arctan((h - e - c)/d)

To maximize θ, we need to find the horizontal distance d that makes the derivative of θ with respect to d equal to zero.

Taking the derivative of θ with respect to d, we have:
d(θ)/dd = -d(h - e - c)/(d^2 + (h - e - c)^2)

Setting this derivative equal to zero and solving for d will give us the optimal value.

After finding the critical points or checking the endpoints (as the case may require), we can determine 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.