Answer:

To maximize the sum of the areas of the rectangles, we want to maximize the product of the length and width of each rectangle. We can use the information given about the sum of the squares of the lengths and widths to solve for the lengths and widths of the rectangles.

Let the lengths of the rectangles be x1, x2, x3, x4, x5, and x6, and let the widths be y1, y2, y3, y4, y5, and y6. Then we have:

x1^2 + x2^2 + x3^2 + x4^2 + x5^2 + x6^2 = 40

y1^2 + y2^2 + y3^2 + y4^2 + y5^2 + y6^2 = 20

To simplify the problem, we can assume that the rectangles are all squares, since squares have the largest area for a given perimeter. So let x1 = x2 = x3 = x4 = x5 = x6 = a, and y1 = y2 = y3 = y4 = y5 = y6 = b.

Then we have:

6a^2 = 40

6b^2 = 20

Solving for a and b, we get:

a^2 = 40/6 = 20/3

b^2 = 20/6 = 10/3

The sum of the areas of the rectangles is:

A = 6ab = 6√(a^2 * b^2) = 6√[(20/3)*(10/3)] = 40√2/3

Therefore, the largest possible value of the sum of the areas of the rectangles is 40√2/3.