Alex's question at Yahoo Answers regarding maximizing viewing angle

[MarkFL]

Here is the question:

How far from the screen should you stand to maximize your viewing angle?


An auditorium with a flat floor has a large screen on one wall. The lower edge of the screen is 3 ft above your eye level.

Optimization problem

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Alex,

First let's draw a diagram:

View attachment 1638

\(\displaystyle s\) is the vertical height of the screen in feet.

\(\displaystyle h\) is the vertical distance from the bottom of the screen to eye level in feet.

\(\displaystyle \theta\) is the viewing angle, which we wish to maximize.

We see that using the definition of the tangent function, we may write:

(1) \(\displaystyle \tan(\beta)=\frac{h}{x}\)

(2) \(\displaystyle \tan(\beta+\theta)=\frac{h+s}{x}\)

Using the angle-sum identity for tangent, (2) may be expressed as:

\(\displaystyle \frac{\tan(\beta)+\tan(\theta)}{1-\tan(\beta)\tan(\theta)}=\frac{h+s}{x}\)

Using (1), this becomes:

\(\displaystyle \frac{\frac{h}{x}+\tan(\theta)}{1-\frac{h}{x}\tan(\theta)}=\frac{h+s}{x}\)

On the left, multiplying by \(\displaystyle 1=\frac{x}{x}\) we obtain:

\(\displaystyle \frac{h+x\tan(\theta)}{x-h\tan(\theta)}=\frac{h+s}{x}\)

Now we want to solve for $\tan(\theta)$. Cross-multiplying, we find:

\(\displaystyle hx+x^2\tan(\theta)=(h+s)x-h(h+s)\tan(\theta)\)

Adding through by \(\displaystyle h(h+2)\tan(\theta)-hx\) we get:

\(\displaystyle h(h+s)\tan(\theta)+x^2\tan(\theta)=(h+s)x-hx=sx\)

Factoring the left side:

\(\displaystyle \left(x^2+h(h+s) \right)\tan(\theta)=sx\)

Dividing through by \(\displaystyle x^2+h(h+s)\) we obtain:

\(\displaystyle \tan(\theta)=\frac{sx}{x^2+h(h+s)}\)

Now, differentiating with respect to $x$ and equating the result to zero to find the critical value(s), we find:

\(\displaystyle \sec^2(\theta)\frac{d\theta}{dx}=\frac{\left(x^2+h(h+s) \right)s-sx(2x)}{\left(x^2+h(h+s) \right)^2}=\frac{s\left(h(h+s)-x^2 \right)}{\left(x^2+h(h+s) \right)^2}=0\)

Multiply through by \(\displaystyle \cos^2(\theta)\) to get:

\(\displaystyle \frac{d\theta}{dx}=\frac{s\cos^2(\theta)\left(h(h+s)-x^2 \right)}{\left(x^2+h(h+s) \right)^2}=0\)

Since we must have \(\displaystyle \theta<\frac{\pi}{2}\), our only critical value comes from:

\(\displaystyle h(h+s)-x^2=0\)

Since \(\displaystyle 0<x\) we take the positive root:

\(\displaystyle x=\sqrt{h(h+s)}\)

Using the first derivative test, we see:

\(\displaystyle \left.\frac{d\theta}{dx} \right|_{x=\sqrt{\frac{h}{2}(h+s)}}=\frac{s\cos^2(\theta)\left(h(h+s)-\frac{h}{2}(h+s) \right)}{\left(\frac{h}{2}(h+s)+h(h+s) \right)^2}=\frac{s\cos^2(\theta)\left(\frac{h}{2}(h+s) \right)}{\left(\frac{3h}{2}(h+s) \right)^2}>0\)

\(\displaystyle \left.\frac{d\theta}{dx} \right|_{x=\sqrt{2h(h+s)}}= \frac{s\cos^2(\theta)\left(h(h+s)-2h(h+s) \right)}{\left(2h(h+s)+h(h+s) \right)^2}= -\frac{s\cos^2(\theta)\left(h(h+s) \right)}{\left(3h(h+s) \right)^2} <0\)

Thus, we conclude the critical value is at a maximum for $\theta$.

Now, using the given value $h=3\text{ ft}$, we find the distance $x$ from the screen that maximizes the viewing angle is given by:

\(\displaystyle x=\sqrt{3(3+s)}\)

 

[Abraxis10]

Thanks for the break down! Adding points of interest in interval notation would be helpful: (0,\infty)

 

[geraldjones]

gerald jones thanks you