Posted by sam on Monday, April 1, 2013 at 10:09pm.
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 crosssectional area is A, determine the dimensions of the window which minimize the perimeter.
h =
w =

calculus  Anonymous, Monday, April 1, 2013 at 11:30pm
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 = (A1.6w)/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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