# calculus

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Consider a window the shape of which is a rectangle of height h surmounted by a triangle having a height T that is 1.1 times the width w of the rectangle

If the cross-sectional area is A, determine the dimensions of the window which minimize the perimeter.

h =
w =

• calculus -

p = w+2h+2s
where (w/2)^2 + (1.1w)^2 = s^2
s = 1.46w

A = wh + 1/2 w * 1.1w
= wh + 1.6w
so, h = (A-1.6w)/w = A/w = 1.6

p = w+2(A/w+1.6) + 2(1.46w)
= w + 2A/w + 3.2 + 2.92w
= 3.92w + 3.2 + A/w

dp/dw = 3.92 - A/w^2
dp/dw = 0 when 3.92w^2 = A, or
w = A/1.98
h = A/1.80

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