Question 305310
A rain gutter is made from sheets of aluminum that are 20 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 13 square inches. 
Show that there are two different solutions to the problem.
 Round to the nearest tenth of an inch. 
:
Let x = the depth of the gutter
Let y = the width of the gutter
:
This may illustrate an end view of the gutter: x|_y_|x
then
2x + y = 20 in; the width of the sheets of aluminum
y = (20-2x); use this form for substitution
and
x*y = 13 sq/in; (the cross sectional area of the end of the gutter)
:
Using the area equation replace y with (20-2x)
x*(20-2x) = 13
a quadratic equation
-2x^2 + 20x - 13 = 0
Solve for x (depth) using the quadratic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
in this equation a=-2; b=20; c=-13
{{{x = (-20 +- sqrt(20^2 - 4*-2*-13 ))/(2*-2) }}}
:
{{{x = (-20 +- sqrt(400 - 104))/(-4) }}}
:
{{{x = (-20 +- sqrt(296))/(-4) }}}
two solutions
{{{x = (-20 + 17.2)/(-4) }}}
x = {{{(-2.8)/(-4)}}}
x = +.7 inches
and
{{{x = (-20 - 17.2)/(-4) }}}
x = {{{(-37.2)/(-4)}}}
x = +9.3 inches 
:
If the depth is .7 inches
20 - 2(.7) = 18.6 in wide; check the area: .7*18.6 ~ 13 sq/in
:
and if the depth is 9.3 in
20 - 2(9.3) = 1.4 in wide; check the area: 9.3*1.4 ~ 13 sq/in also