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# Help idk if I’m doing this right

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4 any help would be appreciated thank You

May 16, 2019

#1
+2

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   May 16, 2019
#2
+3

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) $$\begin{array}{|rclrcl|} \hline \mathbf{\tan(\theta+\alpha)} = \mathbf{ \dfrac{30}{x} }&& \mathbf{\tan(\alpha)} = \mathbf{\dfrac{10}{x}} \\\\ \tan((\theta+\alpha)-\alpha)&=& \dfrac{\tan(\theta+\alpha)-\tan(\alpha)}{1+\tan(\theta+\alpha)\tan(\alpha)} \\\\ &=& \dfrac{\dfrac{30}{x}-\dfrac{10}{x}}{1+\dfrac{30}{x}\cdot \dfrac{10}{x}} \\\\ \mathbf{\tan(\theta)} &=& \mathbf{\dfrac{20x}{x^2+300}} \quad | \quad \text{maximize, take the derivative and set to 0 } \\\\ \tan'(\theta) &=& \dfrac{20x}{x^2+300}\left(\dfrac{20}{20x}-\dfrac{2x}{x^2+300} \right) =0 \\\\ & & \dfrac{1}{x}-\dfrac{2x}{x^2+300} =0 \\\\ && \begin{array}{rclrcl} \dfrac{1}{x} &=& \dfrac{2x}{x^2+300} \\\\ 2x^2&=&x^2+300 \\ x^2&=&300 \\ x &=& \sqrt{300} \\ \mathbf{x} &=& \mathbf{17.3205080757\ feet} \\ \end{array}\\ \hline \end{array}$$ May 17, 2019