1. ## Word Problem. Help please!

A steel pipe is being carried down a hallway 9 ft. wide. At the end of the hall there is a right-angled turn into a narrower hallway 6 ft. wide.

a.) Show that the length of the pipe in the figure is modeled by the function.

L(theta) = 9 csc (theta) + 6 sec (theta)

b.) Graph the function L for 0< (theta)< pi/2.

c.) Find the minimum value of the function L.

d.) Explain why the value of L you found in part (c.) is the length of the longest pipe that can be carried around the corner.

2. 1) We can't see the figure.

2) We are not going to do your homework for you. What have YOU done on this?

3. ok... I am sorry, I don't understand how to go about figuring out the problem. I would like someone to help me so that I can understand this because I have a test next week. I am in college, not high school. Homework doesn't count anymore.

There is a website, if you scroll down to problem 20., thats what the problem in my book looks like.

5. Hello, aligator0207!

Here's part (a).

A steel pipe is being carried down a hallway 9 ft. wide.
At the end of the hall there is a right-angled turn into a narrower hallway 6 ft. wide.

a.) Show that the length of the pipe in the figure is modeled by the function:

. . . . . $\displaystyle L(\theta) \:=\: 9\csc\theta + 6\sec\theta$
Code:
                                      C
*-------------------------------o------
|                           * θ :
|                       *       :6
|              B    *           :
|               o---------------o------
|           * θ |               E
| θ     *       |
|   *           |
A o - - - - - - - o D
|       9       |
|               |

The pipe is: .$\displaystyle L \:=\:AC \:=\:AB+BC$

$\displaystyle AD = 9,\;CE = 6$

Note that: .$\displaystyle \angle ABD \,=\,\angle BCE \,=\,\theta$

In right triangle $\displaystyle BDA\!:\;\;\csc\theta \:=\:\frac{AB}{9} \quad\Rightarrow\quad AB \:=\:9\csc\theta$

In right triangle $\displaystyle CEB\!:\;\;\sec\theta \:=\:\frac{BC}{6} \quad\Rightarrow\quad BC \:=\:6\sec\theta$

Therefore: .$\displaystyle L \;=\;9\csc\theta + 6\sec\theta$

6. Thank you, I understand that more clearly now . Now I think that I can get the next 3 problems. I'll let you know what I get and hopefully I will get them correct!