Call "theta" angle B and "alpha" angle A.....call their sum angle C
tan C = 30/x tan A =10/x
We wish to maximize this
arctan ( B + A) - arctan (A) = arctan (C) - arctan (A)
arctan (C) = 30/x arctan (A) = 10/x
So we have
arctan (30/x) - arctan (10/x) take the derivative of each and set to 0
-30/ [ x^2 +900] - - 10/ [ x^2 + 100] = 0
-30 /x^2 + 900] + 10/{x^2 + 100] = 0
10/ [ x^2 + 100] = 30/ [ x^2 + 900 ] cross-multiply
10 [ x^2 + 900 ] = 30 [ x^2 + 100]
10x^2 + 9000 = 30x^2 + 3000 rearrange as
20x^2 - 6000 = 0
x^2 - 300 = 0
x^2 = 300
x = 10sqrt(3) ≈ 17.32 ft
You are going to see a movie in the theater.
The screen is 20 feet high and 10 feet above the ground.
How far away from the wall should you sit
so hat your viewing angle of the screen is large as possible? (See diagram)
\(\begin{array}{|rclrcl|} \hline \mathbf{\tan(\theta+\alpha)} = \mathbf{ \dfrac{30}{x} }&& \mathbf{\tan(\alpha)} = \mathbf{\dfrac{10}{x}} \\\\ \tan((\theta+\alpha)-\alpha)&=& \dfrac{\tan(\theta+\alpha)-\tan(\alpha)}{1+\tan(\theta+\alpha)\tan(\alpha)} \\\\ &=& \dfrac{\dfrac{30}{x}-\dfrac{10}{x}}{1+\dfrac{30}{x}\cdot \dfrac{10}{x}} \\\\ \mathbf{\tan(\theta)} &=& \mathbf{\dfrac{20x}{x^2+300}} \quad | \quad \text{maximize, take the derivative and set to 0 } \\\\ \tan'(\theta) &=& \dfrac{20x}{x^2+300}\left(\dfrac{20}{20x}-\dfrac{2x}{x^2+300} \right) =0 \\\\ & & \dfrac{1}{x}-\dfrac{2x}{x^2+300} =0 \\\\ && \begin{array}{rclrcl} \dfrac{1}{x} &=& \dfrac{2x}{x^2+300} \\\\ 2x^2&=&x^2+300 \\ x^2&=&300 \\ x &=& \sqrt{300} \\ \mathbf{x} &=& \mathbf{17.3205080757\ feet} \\ \end{array}\\ \hline \end{array}\)