# Thread: Calculus Maximizing and Integrals

1. ## Calculus Maximizing and Integrals

I have a question, because I'm doing this project with my partner for my Calculus class, and we're unsure of how to start and exactly what to do.

Here's what it states:

1. A sign on top of a 50 foot tall building is 10 feet tall. The company president asked you to find the best view of the sign, with pedestrians standing currently about 50 feet and are trampling the company's lawn because the sidewalk is 60 feet away from the building. You expressed a concern, and stated that the best view of the sign should be from 60 feet (on the sidewalk), instead of in the street. The sign must be replaced, so you are to determine a new size for the sign, so that the viewers are safely on the sidewalk 60 feet away from the building. The angle between the lines of sight from the observer's eye to the top and bottom of the sign is a measure of the observer's view of the sign, and best view occurs when the angle is the biggest; the observer's eyes are 5 feet from the ground (observer is 5' 3" tall). Determine the ideal viewing distance from the building when the sign is 10ft, the ideal distance is 49.7 ft. Determine the new sign height when the observer is 60 feet away.

Also, the company is considering putting a bench 35 feet away from the base of the building, so they should also know how tall the sign should be for the best view of the new sign at this distance too.

Any help is greatly appreciated. Thank you.

2. Hello, jjdemick!

Here's the first part . . .

1. A sign on top of a 50-foot tall building is 10 feet tall.
The company president asked you to find the best view of the sign,
with pedestrians standing currently about 50 feet and are trampling
the company's lawn because the sidewalk is 60 feet away from the building.

The angle between the lines of sight from the observer's eye to the top
and bottom of the sign is the measure of the observer's view of the sign,
and best view occurs when the angle is the biggest.
The observer's eyes are 5 feet from the ground (observer is 5' 3" tall).

(a) Determine that, when the sign is 10 ft, the ideal distance is 49.7 ft.
Code:
    A *
| *
10|   *
|     *
B *       *
|   *   θ *
45|       *   *
|        α  * *
C * - - - - - - - * P
5|       x       |5
D * - - - - - - - * Q
x

The flagpole is: $AB = 10$
The building is: $BD = 50$

The observer is: $PQ = 5$.
. . Hence: $CD = 5,\:BC = 45$
The observer's distance from the building is: $DQ = x = CP$

Let $\theta = \angle APB,\;\alpha = \angle BPC$

In right triangle $BCP\!:\;\tan\alpha \,=\,\frac{45}{x}$ .[1]

In right triangle $ACP\!:\;\tan(\theta + \alpha) \:=\:\frac{55}{x} \quad\Rightarrow\quad \frac{\tan\theta + \tan\alpha}{1 - \tan\alpha\tan\theta} \:=\:\frac{55}{x}$

Substitute [1]: . $\frac{\tan\theta + \frac{45}{x}}{1 - \frac{45}{x}\tan\theta} \:=\:\frac{55}{x}$

Multiply by $\frac{x}{x}\!:\;\;\frac{x\tan\theta + 45}{x - 45\tan\theta} \:=\:\frac{55}{x} \quad\Rightarrow\quad x^2\tan\theta + 45x \:=\:55x - 2475\tan\theta$

Then: . $x^2\tan\theta + 2475\tan\theta \:=\:10x \quad\Rightarrow\quad (x^2+2475)\tan\theta \:=\:10x$

. . Hence: . $\tan\theta \;=\;\frac{10x}{x^2+2475}$ . . . and we will maximize $\theta.$

Differentiate: . $\sec^2\!\theta\,\frac{d\theta}{dx} \;=\;10\cdot\frac{2475-x^2}{(x^2+2475)^2} \quad\Rightarrow\quad \frac{d\theta}{dx} \;=\;10\cos^2\!\theta\cdot\frac{2475-x^2}{(x^2+2475)^2} \;=\;0$

. . If $\cos\theta = 0$, then: . $\theta \,=\,\pm90^o$ . . . impossible

. . Therefore: . $2475-x^2\:=\:0 \quad\Rightarrow\quad x \:=\:\sqrt{2475} \:\approx\:\boxed{49.7\text{ ft}}$

3. Thank you for the first part, and what to do. Any idea on the second and third part?

4. Originally Posted by jjdemick
Thank you for the first part, and what to do. Any idea on the second and third part?
1. You've been given an extremely detailed start. In fact, 1/3 of your project has been done.

2. Only 18 minutes have passed between Soroban's post and you asking about parts 2 and 3. Hardly enough time to think about what was posted and how it related to part 2 and part 3, I wouldn't have thought.

3. It's a project. At a wild guess I'd suggest your teacher is a bit more interested in the work of you and your partner than someone elses work.

All this adds up to you showing some effort here. What you've done, where you get stick etc.

This thread is closed for 24 hours. When I re-open it you can post what you've done and where you're still stuck. Appropriate targeted guidance can then be given.

5. Originally Posted by mr fantastic
1. You've been given an extremely detailed start. In fact, 1/3 of your project has been done.

2. Only 18 minutes have passed between Soroban's post and you asking about parts 2 and 3. Hardly enough time to think about what was posted and how it related to part 2 and part 3, I wouldn't have thought.

3. It's a project. At a wild guess I'd suggest your teacher is a bit more interested in the work of you and your partner than someone elses work.

All this adds up to you showing some effort here. What you've done, where you get stick etc.

This thread is closed for 24 hours. When I re-open it you can post what you've done and where you're still stuck. Appropriate targeted guidance can then be given.