# Easy Algebra 2 Questions - Help Please

1. Oct 15, 2009

### bleedblue1234

1. The problem statement, all variables and given/known data

A small motel is to be built as shown in the sketch with two long walls y feet long each and 6 short walls x feet long each. The total length of the walls is to be 300 feet.

Let A(x) be the number of sq feet area taken up by the motel. Write the particular equation for A(x). What kind of function is this?
y
-----------------
| | | | | | x
| | | | | |
-----------------

something like that

2. Relevant equations

y = ax^2 + bx + c

3. The attempt at a solution

no idea where to start

Last edited: Oct 15, 2009
2. Oct 15, 2009

### Staff: Mentor

The total length of all of the walls is 300'.
The six short walls define five rectangular regions - count them in your text drawing. What are the dimensions of the width and length of each rectangle?
What is the area of each rectangle?
What is the total area of all five rectangles?

How is y = ax^2 + bx + c a relevant equation?

3. Oct 15, 2009

### bleedblue1234

yes but it asks for a particular function and i get y*6x and this is not quadratic...

4. Oct 15, 2009

### Staff: Mentor

The problem asks for A(x), meaning a function of just x (no y in it). Since you don't have a function of x alone, you don't know whether the one you really want is quadratic or not. How did you get y*6x as the area?

You haven't taken into account that the total length of all of the walls is 300'.

The six short walls define five rectangular regions - count them in your text drawing. What are the dimensions of the width and length of each rectangle?
What is the area of each rectangle?
What is the total area of all five rectangles?

5. Oct 15, 2009

### arithmetix

The quoted equation is called a quadratic, and if your teacher is responsible you'll have notes on how to deal with quadratics. Usually the job is done by formula:

Given a, b, and c

If ax^2+bx+c=0 then:

x=(-b+-(b^2-4ac))/(2a)

6. Oct 15, 2009

### Staff: Mentor

You're waaaaaay ahead of the OP. He/she is still trying to find a formula for the area of the enclosed rectangles. Let's hold off on solving quadratic equations until he/she at least comes up with a function.

7. Oct 15, 2009

### bleedblue1234

The six short walls define five rectangular regions - count them in your text drawing. What are the dimensions of the width and length of each rectangle? w= x and l = y/6
What is the area of each rectangle? x * y/6 or y*6x
What is the total area of all five rectangles? (5)(y*6x) <= 300

haha holy crap i am bad i am sooo sick tonight i cant even think how to add numbers.... i am usually very good at this sort of stuff and have done fine on all of the other models...

8. Oct 16, 2009

### HallsofIvy

Staff Emeritus

9. Oct 16, 2009

### Staff: Mentor

The width is x, yes, but the length of each one is y/5.
The area of each rectangle is x*y/5. Note that your answer x*y/6 is NOT equal to y*6x.
No. The area of each rectangle is x*y/5. There are 5 rectangles, so the total area is 5*x*y/5 = ?
To summarize, A = 5*x*y/5, which you should simplify. This is the area, but it is not yet a function of x alone. To make it a function of x, you need to work in the other information in this problem, that the total length of the walls, which is 300'.

Look at your drawing and add up the lengths of the 2 walls going across and the 6 walls going up and down. This will give you an equation in x and y.

Solve this equation for y, and substitute for y in your area equation.