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August 5, 2015

August 5, 2015

Posted by **sam ** on Monday, April 1, 2013 at 10:09pm.

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

h =

w =

- calculus -
**Anonymous**, Monday, April 1, 2013 at 11:30pmp = 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