Posted by steph on Thursday, November 10, 2011 at 8:28pm.
Look at a cross-section of the gutter.
let the height be x
then the base width = 20-2x , assuming the gutter is open at the top
a) in 20-2x , 0 < x < 10
b) Area = x(20-2x)
c) Area = 20x - 2x^2
= -2(x^2 - 10x + 25 - 25 )
= -2( (x-5)^2 - 25)
= -2(x-5)^2 + 50
d) maximum area is 50 when x = 5
e) x(20-2x) < 40
-2x^2 + 20x < 40
x^2 - 10x + 20 > 0
consider x^2 - 10x + 20 = 0
x = (10 ± √20)/2
= 5 ± √5
so the area is less than 40 for
5-√5 < x < 5+√5
thank you! that helped a lot
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