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# Restaurant Table Problem. I have tried it on my own but am unsure of my answer. Would appreciate if someone could confirm.

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An architect is designing the dining area of a new restaurant. The owner was to use 3 1/2 ft by 3 1/2 ft square tables to accomodate groups of 4 guests and join two or three tables to accomdate bigger groups. The architect must adhere to the following requirements: Tables in the same row should be space at least 4 ft. apart and two rows of tables should be spaced at least 5 ft apart. For tables placed next to walls, a minimum of 2 ft spacing is required between the wall and table.

1. What is the minimum (as close to square shaped as possible) dining area (in square feet) required to accomodate around 100 guests in groups of four?

2. What is the maximum number of guests that can be accommodated using the same area and same number of tables as in the first question if roughly 24% of the tables are reconfigured as bigger tables by joining 2 tables and roughly 12% of the tables are reconfigured by joining three tables?

I have already tried solving this problem, but am unsure if it is correct. I would appreciate it if someone else could walk through their thought process and confirm my answer.

Here is how I did it:

To determine the number of tables necessary to seat 100 people, I have to divide the number of guests by the number of people per table. This number is 4, so 100/4 = 25 tables.If the answer should be as square shaped as possible and there are 25 tables, then there should be 5 tables per row. Now, I must determine the width and length of the square shape I have created. Since there are 5 tables and each table measures 3 1/2 feet, I can multiply this to get 17.5. For the width, there must be a distance of 4 feet, but this will only be present four times. For this reason, I can multiply 4*4 to get 16. To get the distance from the walls, I assumed that each horizontal side had some sort of wall, as the restaurant is probably enclosed. Because of this, I can multiply 2*5, as there are 5 tables in each row and they are all next to a horizontal wall. This is 10. Now, I can add the values up to get 43.5. I used the same process for the length, but used 5*4, instead of 5, as the distance between each row is 5. This total was 47.5. I multiplied the two numbers to get 2066.25 or 2066 square feet of space.

If 24% of the 25 tables are now 2 tables, then there are 24/100 * 25 = 6 sets of 2 tables. To determine the number of people these sets can seat, I came up with a function y = 2x+2, where x is the number of tables and y is the number of people. If there are 2 tables, then they can seat up to 6 people. Four people will sit across from each other, while two sit on the sides. If there are 6 sets of this type of table, then it is simply 6*6, which is 36 more people. To find the number of people that can be seated using three tables, I used the same equation. This time, 2(3) + 2 = 8 people. If 12% of the tables are reconfigured, then 12/100*25 = 3 sets of 3 tables, which requires multiplying 8*3 to get 24 more people. Adding 36 and 24 to the already 100 people seated gets 160 people in total.

I also assumed a few things:

I assumed that both the horizontal row and vertical column were surrounded by walls and added this onto the other calculations. I also assumed that reconfiguration of the tables involved joining more tables onto the current ones, as the problem suggests that more people will be seated in addition to the 100 people that already have seats.

Jan 22, 2022