Question 229709
a rain gutter is made from sheets of aluminum that are 27 inches wide.
 The edges are turned up to form right angles.
 Determine the depth of the gutter that will allow a cross-sectional area of 58 inches. 
There are 2 solutions to this problem.
:
end view y=depth, x=width: 
y|_x_|y
:
2y + x = 27"
x = (27-2y)
:
Cross sectional area
x*y = 58 sq/inches
:
Replace x with (27-2y)
y(27-2y) = 58
-2y^2 + 27y - 58 = 0
:
Use the quadratic formula to find y
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this equation x=y; a=-2; b=27; c=-58
{{{y = (-27 +- sqrt(27^2 - 4*-2*-58 ))/(2*-2) }}}
:
{{{y = (-27 +- sqrt(729 - 464 ))/(-4) }}}
:
{{{y = (-27 +- sqrt(265))/(-4) }}}
Two solutions
{{{y = (-27 - 16.2788)/(-4) }}}
{{{y = (-43.2788)/(-4) }}}
y = +10.82 inches is the depth
and
{{{y = (-27 + 16.2788)/(-4) }}}
{{{y = (-10.72)/(-4) }}}
y = +2.68 inches is the depth
:
Both these solutions satisfy the requirement, the width will vary accordingly
:
Check solution using the last solution
find x
2(2.68) + x = 27
x = 27 - 5.36
x = 21.64 is the width
;
Find the cross sectional area
2.68 * 21.64 = 57.99 ~ 58
:
You can check it using the other solution.