Consider a window the shape of which is a rectangle of?

didn't I already answer this? If you have a problem with the answer, contact me.

Area of rectangle is wh
area of triangle is (1/2)wT
total area is wh + (1/2)wT
T = 0.8w
A = wh + (1/2)w(0.8w)
A = wh + 0.4w²

perimeter P = 2h + w + 2x
where x is the angled side of the top
if you divide the top section via a vertical line, you have two right triangles.
x² = T² + (w/2)²
T = 0.8w
x² = 0.64w² + (w/2)² = 0.89w²
x = 0.9434w

P = 2h + w + 2(0.9434w)
P = 2h + 2.886796w
to minimize the P, we need to get P in terms of one variable.
h = (A/w) – (0.4w²/w)
h = (A/w) – (0.4w)
P = 2h + 2.886796w
P = 2((A/w) – (0.4w)) + 2.886796w
P = (2A/w) – (0.8w) + 2.886796w
P = (2A/w) + 2.086796w

now finally we can differentiate P and set equal to 0 to find min
assuming A is fixed
P' = 0 = 2.086796 – 2A/w²
solve for w
2A/w² = 2.086796
w² = 2A/2.086796 = A(0.9584069)
w = 0.9789826√A

h you can get from h = (A/w) – (0.4w)

edit, I did see one math mistake, correcting now
A lot of math, you should check it.

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