<PRE><B>Question 268741</B>
a 15-inch by 28-inch poster is to be framed using a border of uniform width.
 to make the framing visually appealing, the border should have an area equal to 60% of the area of the poster.
 how wide should the border be?
:
Let x = the width of the border
:
Find the area of the poster:
15 * 28 = 420 sq/in
:
Find the area of the border
.6(420) = 252 sq/in
:
The overall dimensions of poster and border:
(2x+15) by (2x+28)
:
Overall area - poster area = border area
(2x+15)*(2x+28) - 420 = 252
FOIL
(4x^2 + 56x + 30x + 420) - 420 = 252
:
4x^2 + 86x + 420 - 420 = 252
:
4x^2 + 86x - 252 = 0; a quadratic equation
:
Use the quadratic formula to find the positive solution for x
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this equation: a=4; b=86; c=-252
{{{x = (-86 +- sqrt(86^2-4*4*-252 ))/(2*4) }}}
:
{{{x = (-86 +- sqrt(7396 - (-4032) ))/8 }}}
:
{{{x = (-86 +- sqrt(7396 + 4032 ))/8 }}}
:
{{{x = (-86 +- sqrt(11428 ))/8 }}}
:
{{{x = (-86 + 106.9)/8 }}}
:
x = {{{20.9/8}}}
x = 2.61 inches is the width of the border
:
:
You can check solution:
(15+5.22)*(28+5.22) - 420 = should equal 60% of 420</PRE>