# Alex's question at Yahoo! Answers regarding maximizing viewing angle

#### MarkFL

Staff member
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

Staff member
Hello Alex,

First let's draw a diagram:

$$\displaystyle s$$ 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

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

#### geraldjones

##### New member
gerald jones thanks you