SOLUTION: I have a rectangle dry erase board that's 48" Wx 36"L I need to fit 35 EQUAL SIZED squares inside it, SUCH THAT EACH SQUARE IS AS LARGE AS POSSIBLE. I need to know the size of

Algebra ->  Rectangles -> SOLUTION: I have a rectangle dry erase board that's 48" Wx 36"L I need to fit 35 EQUAL SIZED squares inside it, SUCH THAT EACH SQUARE IS AS LARGE AS POSSIBLE. I need to know the size of       Log On


   



Question 1078326: I have a rectangle dry erase board that's 48" Wx 36"L I need to fit
35 EQUAL SIZED squares inside it, SUCH THAT EACH SQUARE IS AS LARGE
AS POSSIBLE. I need to know the size of each square.
Thank you

Found 2 solutions by Fombitz, Edwin McCravy:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
The smallest size of the rectangular board will dictate the maximum size of the squares.
So you want to fit 35 squares into 36 inches, so the size of each square (S) should be,
S=36%2F35in

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
I have a rectangle dry erase board that's 48" Wx 36"L I need to fit
35 EQUAL SIZED squares inside it, SUCH THAT EACH SQUARE IS AS LARGE
AS POSSIBLE. I need to know the size of each square.
Thank you
I am quite sure the solution Fombitz posted above is NOT what you want.

Notice that I have added some words in capital letters to your
problem above.  Let me know in the thank-you note form below if
the words I added were not to be assumed.  If we did not make
the assumption that the squares must be EQUAL in size and as 
LARGE AS POSSIBLE, then there would be more than one solution,
in fact maybe infinitely many solutions.

We have to draw 5 squares along the shorter side of the
whiteboard and 7 squares along the longer side to have 35
squares, each as large as possible.

First we determine if we can use the entire shorter 36" side of
the board for the 5 squares.

Then each square would be 36÷5 = 7 1/5 inches on a side.
But then we would have to fit 7 of those across the longer 48
side.  But (7 1/5)×7 = 50 2/5", which is too long for 7 squares
to fit on the 48" side. So we cannot use the entire shorter
side for the 5 squares.

That means we must use the entire longer 48" side for the
7 squares. So to fit 7 squares across the longer 48" side,
each square must be 48÷7 = 48/7 = 6%266%2F7".


That's the answer.  Each square would have sides of 
6%266%2F7 inches.  Below is how the whiteboard would look
after drawing the 35 squares each 6%266%2F76%266%2F7".

 

Edwin