Real-World question i can't get my head around.

[JoshMaths]

I am going to feel silly when you guys give me the answer in 2 seconds but here we go.

Imagine there are 16 rectangular pieces of paper with length 2 and width 1.

You want to organise them in a big square or make them fit the smallest area possible.

How many columns do you have and how many rows?

If you need a pitiful attempt then i would say...
Let x be the total length and y be the total width then x=2y and r*c = 16 where r = rows and c = columns and we want to find min xy = min 3y²

given x = 2r and y = c then 3y² = 32 yet this minima gives y = 0 obviously, so i am stuck.

Yes i know this is year 8 maths, yes i am in University, any help much appreciative.

 

[mathman]

No matter how you arrange them the area will be 32. Since 32 is not a perfect square you can't make a square. You can make rectangles of any shape (1 x 16, 2 x 8, 4 x 4) where either direction can be all 1's or all 2's. It is also possible to mix 1's and 2's.

 

[JoshMaths]

Thanks, one constraint i might have forgot to mention is that all the pages must be vertical so you can read them.

 

[mfb]

mathman's solution can be done with vertical pages. Another nice arrangement could be 3 vertical, 6 horizontal. It requires 36 tiles instead of 32, but it gives a nice 6x6-pattern (with 4 empty spots) - the smallest possible square.

 

[CellCoree]

I can understand how this problem may seem challenging at first. However, it is important to approach it systematically and logically.

First, let's consider the dimensions of the square that we want to make with the 16 rectangular pieces of paper. Since each piece has a length of 2 and a width of 1, the total length of the square would be 2*4=8 and the total width would be 1*4=4.

Next, we need to determine the number of rows and columns that would give us the smallest area for this square. We can use the formula A=l*w to calculate the area, where A is the area, l is the length, and w is the width.

Since we want to minimize the area, we can set up the following equation: A=8r*4c=32rc.

We also know that there are 16 rectangular pieces, so we can set up another equation: r*c=16.

Now, we can substitute the value of r*c from the second equation into the first equation to get: A=32(16)=512.

To minimize the area, we need to find the factors of 512 that give us the smallest value. The factors of 512 are 1, 2, 4, 8, 16, 32, 64, 128, 256, and 512.

Since we want the smallest value, we can start by trying the smallest factors, which are 1 and 2. When we plug in these values for r and c, we get A=32(1)(2)=64. This is the smallest possible value for the area, which means that we need to have 1 row and 2 columns to make the smallest square with the 16 rectangular pieces of paper.

I hope this explanation helps you understand the solution to this problem. Don't feel silly for not immediately knowing the answer, as it takes practice and logical thinking to solve mathematical problems. Keep up the good work in your studies!