Questions: An auditorium with a flat floor has a large screen on one wall. The lower edge of the screen is 3 ft above eye level and the upper edge of the screen is 11 ft above eye level (see figure). How far from the screen should you stand to maximize your viewing angle? You should stand ft from the screen to maximize your viewing angle. (Type an exact answer, using radicals as needed.)

An auditorium with a flat floor has a large screen on one wall. The lower edge of the screen is 3 ft above eye level and the upper edge of the screen is 11 ft above eye level (see figure). How far from the screen should you stand to maximize your viewing angle?

You should stand ft from the screen to maximize your viewing angle.
(Type an exact answer, using radicals as needed.)

Transcript text: An auditorium with a flat floor has a large screen on one wall. The lower edge of the screen is 3 ft above eye level and the upper edge of the screen is 11 ft above eye level (see figure). How far from the screen should you stand to maximize your viewing angle? You should stand $\square$ ft from the screen to maximize your viewing angle. (Type an exact answer, using radicals as needed.)

Solution

Solution Steps

Step 1: Define the Problem

We need to find the distance $ x $ from the screen to maximize the viewing angle $ \theta $. The lower edge of the screen is 3 ft above eye level, and the upper edge is 11 ft above eye level.

Step 2: Set Up the Viewing Angle

The viewing angle $ \theta $ is the angle between the lines of sight to the top and bottom of the screen. We can express $ \theta $ as the difference between two angles: \[ \theta = \alpha - \beta \] where $ \alpha $ is the angle to the top of the screen and $ \beta $ is the angle to the bottom of the screen.

Step 3: Express Angles in Terms of $ x $

Using trigonometry, we can express $ \alpha $ and $ \beta $ in terms of $ x $: \[ \tan(\alpha) = \frac{11}{x} \] \[ \tan(\beta) = \frac{3}{x} \]

Step 4: Use the Difference of Angles Formula

The tangent of the difference of two angles is given by: \[ \tan(\theta) = \tan(\alpha - \beta) = \frac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha)\tan(\beta)} \] Substitute the expressions for $ \tan(\alpha) $ and $ \tan(\beta) $: \[ \tan(\theta) = \frac{\frac{11}{x} - \frac{3}{x}}{1 + \frac{11}{x} \cdot \frac{3}{x}} = \frac{\frac{8}{x}}{1 + \frac{33}{x^2}} = \frac{8/x}{(x^2 + 33)/x^2} = \frac{8x}{x^2 + 33} \]

Step 5: Maximize $ \tan(\theta) $

To maximize $ \theta $, we need to maximize $ \tan(\theta) $: \[ \frac{d}{dx} \left( \frac{8x}{x^2 + 33} \right) = 0 \] Use the quotient rule for differentiation: \[ \frac{d}{dx} \left( \frac{8x}{x^2 + 33} \right) = \frac{(8)(x^2 + 33) - (8x)(2x)}{(x^2 + 33)^2} = \frac{8x^2 + 264 - 16x^2}{(x^2 + 33)^2} = \frac{-8x^2 + 264}{(x^2 + 33)^2} \] Set the numerator equal to zero to find the critical points: \[ -8x^2 + 264 = 0 \] \[ 8x^2 = 264 \] \[ x^2 = 33 \] \[ x = \sqrt{33} \]