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 = ".


That's the answer.  Each square would have sides of 
 inches.  Below is how the whiteboard would look
after drawing the 35 squares each "×".

 

Edwin