The pentagon is to be constructed by placing an equilateral triangle on a rectangle and the dimensions of the two determined to maximize the area of the pentagon. Let the rectangle have the dimensions L x W and the equilateral triangle is placed on the side with length L.

As the perimeter is 30: `3*L + 2*W = 30`

The area of the pentagon is A = `L*W + (sqrt 3/4)*L^2`

=> `L*(30 - 3L)/2 + (sqrt 3/4)*L^2`

=> `15L - 3L^2/2 + (sqrt 3/4)*L^2`

=> `15L + (sqrt 3/4 - 3/2)*L^2`

To maximize the area solve `(dA)/(dL) = 0`

=> `15 + 2*L*(sqrt 3/4 - 3/2) = 0`

=> `L = -15/(sqrt 3/2 - 3)`

=> `L = 15/(3 - sqrt 3/2)`

=> `L = 30/(6 - sqrt 3)`

The width of the rectangle is `W = (30 - 3*30/(6 - sqrt 3))/2`

=> W = `15 - 45/(6 - sqrt 3)`

To maximize the area of the pentagon the dimensions of the rectangle are `30/(6 - sqrt 3)` x `15 - 45/(6 - sqrt 3)` and the equilateral triangle has sides equal to `30/(6 - sqrt 3)`