How Do You Calculate the Area of a Motel Given Wall Lengths?

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    Algebra Algebra 2
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Homework Help Overview

The problem involves calculating the area of a motel based on given wall lengths, specifically with two long walls and six short walls, while adhering to a total wall length constraint of 300 feet. The original poster seeks to express the area as a function of one variable, A(x).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the dimensions of the motel's rectangular regions defined by the short walls and question how to derive the area function A(x) from the given parameters. There is confusion regarding the relevance of the quadratic equation provided and how it relates to the area calculation.

Discussion Status

The discussion is ongoing, with participants providing guidance on how to approach the problem. Some participants emphasize the need to derive a function of x alone for the area, while others reiterate the importance of understanding the dimensions of the rectangles involved.

Contextual Notes

There is a noted constraint regarding the total length of the walls being 300 feet, which is central to forming the equations needed to express the area as a function of x. Participants are also addressing potential misunderstandings in the calculations of area based on the motel's layout.

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Homework Statement



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

Homework Equations



y = ax^2 + bx + c

The Attempt at a Solution



no idea where to start
 
Last edited:
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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?
 
Mark44 said:
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?

yes but it asks for a particular function and i get y*6x and this is not quadratic...
 
bleedblue1234 said:
yes but it asks for a particular function and i get y*6x and this is not quadratic...
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'.

I'll ask my questions again.
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?

I'll keep asking them until you answer them.
 
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)
 
arithmetix said:
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)

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.
 
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 can't 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...
 
arithmetix said:
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)
Please read questions carefully before you respond. The original post asked only for the area as a quadratic equation. Nothing is said about solving the equation.
 
bleedblue1234 said:
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
The width is x, yes, but the length of each one is y/5.
bleedblue1234 said:
What is the area of each rectangle? x * y/6 or y*6x
The area of each rectangle is x*y/5. Note that your answer x*y/6 is NOT equal to y*6x.
bleedblue1234 said:
What is the total area of all five rectangles? (5)(y*6x) <= 300
No. The area of each rectangle is x*y/5. There are 5 rectangles, so the total area is 5*x*y/5 = ?
bleedblue1234 said:
haha holy crap i am bad i am sooo sick tonight i can't 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...

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.
 

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