best viewing angle...

[bluester105]

in a certain movie theater, the bottom edge of the 15 foot tall screen at the front is 5 feet above eye level. suppose we wish to sit so that the angle theta is as large as possible. how far should we sit from the screen?

so i made a diagram of what i am supposed to be looking for....
...|\
...|.\
15|...\ ...t=theta.
...|\....\
...|..\.t./\
...|....\/..\
5.|.......\..\
...|..........\.\
...|.............\\
...|_____x____| (sorry it looks better on paper....)


okay so what i dont understand is how i am supposed to incorperate x and theta into the same problem..i hope what i said makes sense...thanks for the help..

 

[renegade05]

Paint man!

Is this your problem ?

You want to maximize theta in terms of x?

I would start by writing some trig relationships down. See if you can get an equation "theta = something".

HINT: THINK ARCTAN

 

[soroban]

Hello, bluester105!

In a certain movie theater, the bottom edge of the 15-foot tall screen at the front is 5 feet above eye level.
Suppose we wish to sit so that angle theta is as large as possible.
How far should we sit from the screen?

      A *
        | *
        |   *
     15 |     *
        |       *
        |         *
      B *           *
        |     *     θ *
      5 |         α *   *
        *-----------------*
        C - - -  x  - - - P

The screen is $\displaystyle AB = 15.$
It is 5 feet above eye level: $\displaystyle BC = 5.$
We are sitting at point $\displaystyle P\!:\:x = CP.$
Let $\displaystyle \theta = \angle APB,\:\alpha = \angle BPC.$

We have: .$\displaystyle \tan(\theta + \alpha) \:=\:\dfrac{20}{x} \quad\Rightarrow\quad \theta + \alpha \:=\:\tan^{\text{-}1}\!\left(\dfrac{20}{x}\right)$
Hence: .$\displaystyle \theta \;=\;\tan^{\text{-}1}\!\left(\dfrac{20}{x}\right) - \alpha$ .[1]

We have: .$\displaystyle \tan\alpha \:=\:\dfrac{5}{x} \quad\Rightarrow\quad \alpha \:=\:\tan^{\text{-}1}\!\left(\dfrac{5}{x}\right)$

Sustitute into [1]: .$\displaystyle \theta \;=\;\tan^{\text{-}1}\!\left(\dfrac{20}{x}\right) - \tan^{\text{-}1}\!\left(\dfrac{5}{x}\right)$


Differentiate and equate to zero:

. . $\displaystyle \dfrac{d\theta}{dx} \;=\;\dfrac{-\frac{20}{x^2}}{1 + \left(\frac{20}{x}\right)^2} - \dfrac{-\frac{5}{x^2}}{1 + \left(\frac{5}{x}\right)^2} \;=\;0$

. . . . . . . . .$\displaystyle \dfrac{-20}{x^2+400} + \dfrac{5}{x^2+25} \;=\;0 \quad\Rightarrow\quad \dfrac{20}{x^2+400} \;=\;\dfrac{5}{x^2+25}$

. . . . . . . $\displaystyle 20(x^2 + 25) \;=\;5(x^2+400) \quad\Rightarrow\quad 20x^2 + 500 \;=\;5x^2 + 2000$

. . . . . . . $\displaystyle 15x^2 \;=\;1500 \quad\Rightarrow\quad x^2 \;=\;100$

$\displaystyle \text{Therefore: }\:x \:=\:10\text{ feet.}$