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Finding the Dimensions of a Room


Date: 12/5/95 at 20:54:44
From: Anonymous
Subject: Fariba's Nightclub - High School word problem

My daughter's Algebra I teacher gave her students this word problem 
as a challenge:

Mrs. Fariba Suarez has decided to use the profits from her 
swimwear company to open a night club.  After studying much 
statistical data, Fariba surmises that this club should seat 
a minimum of 1152 people at tables.  The tables will be 3 x 6 
feet, and each one will seat 6 customers.  Fire codes require 
that all tables be at least 3 feet from the wall and that there 
must be at least a 3-foot space between the tables.  There must 
be a 6-foot wide walkway through the center of the room parallel 
to the width of the building.

Fariba must now decide on the dimensions of the building.  
Assuming the length of the building will be twice the width, 
find the dimensions of the smallest possible building Fariba 
can construct and describe the arrangement of the tables in 
the building.

She has figured out that there must be a minimum of 192 tables by
dividing 6 into 1152.  This allows for 96 tables on each side of 
the room.  

She has figured out that 1 1/2 feet should be added to each side of
the table to make the table dimension 4 1/2 x 7 1/2 feet.  When two
tables are next to each other, this will allow for a 3-foot space 
that is required by the fire code.

She has figured out that a 1 1/2 foot space must be added to each wall 
to allow for the 3 foot space required by the fire code.  This space
will also absorb the 1 1/2 allowed with each table.

A 3-foot dimension must be added to the length to allow for the 
6-foot walkway through the center of the room that is parallel to the 
width of the room.

She needs help in placing all of this information into an equation.

Thanks, Kathy 


Date: 5/31/96 at 11:22:32
From: Doctor Alain
Subject: Re: Fariba's Nightclub - High School word problem

Each table needs an area of 6x9 square feet. For each table, imagine
there is an extra 3 feet on the north side and 3 feet on the east
side. Each table takes care of the extra space needed on the north
and east. The tables to the south and to the west will take care
of the space needed to the south and to the west.  This also provides
for the wall space for the north and east walls but not for the
south and west walls; this extra space will be added later on.

Imagine a wall in the center of the walkway. You would have two
half rooms, each containing half the tables, and a 3-foot spacing 
at each wall (6/2 = 3 so 3 feet on each side of the imaginary wall 
also). You need 1152/6 = 192 tables, that is, 96 per half room.
The tables will be in a square area of at least 96x6x9= 5184 square
feet. 

Figure out the lengths of the sides of a square of this area
and whether you can fit in the tables. (For this problem it can
be done but it isn't always so; for instance, if you had a single
table 4 feet by 9 feet or 36 square feet, you couldn't fit it in
a square 6 feet by 6 feet.) 

Once you have found a way to fit the tables in, don't forget to add 
the extra 3 feet along the south and west walls and to double the 
size of the building to get the other side of the walkway.

There are many possible arrangements for the tables. Try to find
as many as you can.

-Doctor Alain,  The Math Forum

    
Associated Topics:
High School Basic Algebra

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