Mary's square tiles

Let x be the width in tiles of the red rectangle and y be the height of rectangle. We notice that the number of blue tiles used will be 2x + 2y +4. If this is 62 then this gives us a very nice equation. Y=29-x. Let f(x) be the area of the red rectangle which is x * y or x(29-x). We examine this graph and notice a maximum at 14.5. since we are using whole tiles, we round to 15. So x is 15 and y is 14. Area is 14 * 15 = 210.

Let the perimeter be 2(x+y)=62 => x+y=31. Now to have the maximum number of red tiles, which corresponds to maximum area, we need 2 numbers that are either equal or as close as possible and add up to 31. This gives us 15 and 16. Now, since we want the are of only the red tiles and not the blue tiles, we remove 1 unit from each number, i.e. 15-1 and 16-1. now we have two numbers that indicate the sides of the rectangle formed by the red tiles. we multiply them to get the answer. 14*15=210

Below is a solution without the use of algebra.

62 blue tiles in total - 4 blue corner tiles = 58 = Perimeter of red rectangle.

Length + Width of red rectangle = 58/2 = 29.

The number of red tiles is maximised (i.e. the red rectangle's area is also maximised) when the length and width of the rectangle are almost equal, in this case we can have the length = 15 and width = 14.

Hence the number of red tiles used = 15*14 = 210.

From the starting of red tiles, in the first row and column no. of tiles = 54(-8 from 62 blue tiles) and each time no. of tiles decrease by 8(each corner by 2) so do it still you can form the rectangle 54+46++38+30+22+14+6=210

here 62 blue tiles which make rectangle at peripheral and inside it red tiles make rectangle, which tiles required have to be measured. First divide 62 tiles at 4 part that is up,down,right ,left to form a rectangle. I consider two horizontal part each contains 16 blue tiles, and two vertical part each contains 15 tiles plus each contain two tiles which we count for horizontal section, that's why Here i am not mention it.Now to draw a red tiles rectangle inside it, we have(16-2=14) tiles slot at horizontal and (17-2=15) tiles slot at vertical. So number of red tiles required is=14*15(14 multiple 15)=210 .

From the starting of red tiles, in the first row and column no. of tiles = 54(-8 from 62 blue tiles) and each time no. of tiles decrease by 8(each corner by 2) so do it still you can form the rectangle 54+46++38+30+22+14+6=210

As you can see on the picture no matter the perimeter of the red rectangle, the number of blue squares is always gonna be greater by 4 digits. So since the maximum number of blue tiles is 62, the maximum perimeter of the inner red rectangle should be 58. --> 2x + 2y = 58 --> xy = number of red tiles So I just used the quadratic formula to find x and y that were integers since we're dealing with tiles and it worked only when xy=210 (which means x= 15 and y= 14). Hence 210 is the answer

Since, the tiles are in square shape and we have to set them in a rectangle shape the number of rows and columns will be more than one. Let x is number of rows of red tiles and y is the number of columns of red tiles, then total number of RED tiles will be (xy). To make a rectangle y should be at least 1 greater than x (we can say y=x+1). Number of BLUE tiles at periphery will be 2(x+2)+2y=62 and solving this we get x+y=62, but we have also y=(x+1) putting this we get x+(x+1)=29 therefore, x=14. and y=15 xy=14*15=210, this is the answer.