A square with sides of length 10 units is shown in the diagram - Junior Cycle Mathematics - Question 7 - 2011

To find the total area of the two shaded squares, let's denote the side of the left square as A. The area of the left square is given by A², while the area of the right square, which has a side of length (10 - A), is given by (10 - A)². Consequently, the total area, T, can be expressed as:

$T = A^2 + (10 - A)^2$

Expanding the right-hand side, we get:

$T = A^2 + (100 - 20A + A^2) = 2A^2 - 20A + 100$

To find the minimum area, we can take the derivative of T with respect to A and set it to zero:

$\frac{dT}{dA} = 4A - 20$

Setting the derivative equal to zero:

$4A - 20 = 0 \Rightarrow A = 5$

To confirm that this is a minimum, we can check the second derivative:

$\frac{d^2T}{dA^2} = 4$

Since the second derivative is positive, the function has a minimum at A = 5. Now substituting A = 5 into the total area equation:

$T = 2(5)^2 - 20(5) + 100 = 50$

Thus, the minimum possible value of the total area of the two shaded squares is 50.