Calculus help please?

Question

At the outdoor summer movie at Stanley Park, the bottom of the screen is 6 meters tall, and is 2 m above eye level. At what distance x from the base of the screen is the visual angle occupied by the screen as large as possible?

(HINT: One approach is to define θ as the angle to the top of the screen and ϕ as the angle to the bottom of the screen and maximize their difference.)

Please explain your answers so I may understand these concepts.

Saurabh Dubey · Answer

...tanθ =(8/x) ,  tanф =(2/x)

tan(θ-ф) =(tanθ - tanф)/(1+tanθ.tanф)

...=((8/x)-(2/x))/(1+(8/x).(2/x))

...=(6x/(x²+16))

Let visual angle=(θ-ф)=u
 
 ...tanu  =(6x/(x²+16))______(1)

Now we have to maximize 'u' so we will find 
out (du/dx) and equate it to zero

Differentiating (1) w.r.t  x

...sec²u (du/dx)=((x²+16).6 -(2x.6x))/(x²+16)²

...du/dx = ((96 - 6x²)/(x²+16)²)cos²u_______

We know that (sec²u =tan²u +1)

...sec²u = (6x/(x²+16))² +1

... = (36x²+x⁴+256+32x²)/(x²+16)²

....= (x⁴+68x²+256)/(x²+16)²

...cos²u = (x²+16)²/(x⁴+68x²+256)_______

so,

du/dx = ((96 - 6x²)/(x²+16)²) .

((x²+16)²/(x⁴+68x²+256))

...du/dx = (96-6x²)/(x⁴+68x²+256)

For maximisation  ,

...du/dx =0

...(96-6x²)/(x⁴+68x²+256) =0

...96-6x² =0

...x²=16

...x=± 4

Draw sign scheme

.....(+)......-4.......(-).......4.......(+).......

At x = -4  sign changes from (+) to (-)

so , at x = - 4 ( i.e to the left of base of the screen at a distance of 4 unit) visual angle will be largest .

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Calculus help please?

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