Question 1107733
A sheet of metal, 60cm wide and of indefinite length is bent to form an open rectangular gutter.
 What height of the gutter will give the maximum cross-sectional area?
 What height will give a cross-sectional area of 300cm^2?
:
A rough picture of the end of the gutter
 h|_w_|h
h = the height of the gutter,
w = the width
:
2h + w = 60
w = (60-2h), we can use this form for substitution
:
Cross sectional area = width * height
f(h) = the cross sectional area, replace w with (60-2h)
f(h) = h(60-2h)
f(h) = 60h - 2h^2
max h occurs at the axis of symmetry a = -b/(2a), where a = -2, b = 60
h = {{{(-60)/(2*-2)}}}
h = +15 cm is the height for max area
;
"What height will give a cross-sectional area of 300cm^2?"
60h - 2h^2 = 300
Arrange as a quadratic equation
-2h^2 + 60h - 300 = 0
simplify, divide by -2
 h^2 - 30h + 150 = 0
use the quadratic formula; a=1, b=-30, c=150
I got two solutions
h = 6.34 cm, then w = 60 - 2(6.34) = 47.32 cm is the width
and
h = 23.66 cm, then w = 60 - 2(23.66) = 12.68 cm is the width
:
:
Check this, find the cross section area with these value
6.34 * 47.32 = 300 sq cm
and
23.66 * 12.68 = 330 sq cm