(Diagam not to scale)

We define θ1 and θ2 as shown in the diagram such that:

`θ = θ_2 - θ_1`.

Let x be the distance from directly under the screen to the observer. To maximise `θ,` we will need to find

`dy/dx`

and then set it to 0.

We note that

`tan\ theta_1=8.5/x`, and

`tan\ theta_2=10.9/x`

This gives:

`theta_1=tan^-1(8.5)/x ` and

`theta_2=tan^(-1) 10.9/x`

Now since `θ = θ_2 - θ_1`,

`theta=tan^-1 10.9/x-tan^-1 8.5/x`

We have a function of a function in each term.

Now, in the first term, if we let

`u=10.9/x=10.9x^-1`

then

`(du)/(dx)=d/(dx)(10.9x^-1)`

`=-10.9x^-2`

`=-10.9/x^2`

Similarly for the second term, we will have:

`d/(dx)(8.5x^-1)`

`=-8.5x^-2`

`=-8.5/x^2`

So we have:

`(d theta)/(dx)=1/(1+((10.9)/x)^2)(-10.9/x^2)` `-1/(1+((8.5)/x)^2)(-8.5/x^2)`

`=1/x^2((-10.9)/(1+(10.9^2)/(x^2))+8.5/(1+(8.5^2)/(x^2)))`

Next, we multiply the x2 in the denominator (bottom) of the first fraction by the denominators of the 2 fractions in brackets, giving:

`=(-10.9)/(x^2+10.9^2)+8.5/(x^2+8.5^2)`

`=(-10.9(x^2+8.5^2)+8.5(x^2+10.9^2))/((x^2+10.9^2)(x^2+8.5^2))`

`=(-2.4x^2+222.36)/((x^2+10.9^2)(x^2+8.5^2))`

To find when this equals `0`, we need only determine when the numerator (the top) is `0`.

That is

`-2.4x^2 + 222.36 = 0`

This occurs when `x = 9.63` (we take positive case only)

So the observer must be `9.63\ "m"` from directly below the screen to get the best view.

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