Let the rectangle have width x and height y. Since you specifically say "the six pieces", I presume there is a piece of frame between the triangular and rectangular pieces. The length of the frame is 4x+ 2y (4x because the base and top of the rectangle have length x and the triangular top has two sides of length x). Then the area of the rectangular part of the window is xy and the area of the triangular part is $\frac{\sqrt{3}}{4}x^2$. The entire area is $xy+ \frac{\sqrt{3}}{4}x^2$. Minimize that subject to the conditions $4x+ 2y\le 6$, $0 \le x\le 1$, and $0\le y\le 3$. (I am assuming that "The window must fit inside a space 1 m wide and 3 m wide" should have been "The window must fit inside a space 1 m wide and 3 m high".)
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