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any help would be appreciated thank You 

 May 16, 2019
 #1
avatar+105277 
+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

 

 

cool cool cool

 May 16, 2019
 #2
avatar+23521 
+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)

 

 

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

 

laugh

 May 17, 2019

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