<PRE><B>Question 4462</B>
Begin by drawing a rectangle whose width is x, and whose length is 2x.  Add another rectangle outside this rectangle that extends 4 feet in each direction from the inner rectangle.  The width of the outer rectangle is x+8, and the length of the outer rectangle is 2x+8, since you have to add 4 feet TWICE in each direction.


The area of the inner rectangle is LW = x(2x) or {{{2x^2}}}.
The area of the outer rectangle is LW = (x+8)(2x+8) or {{{2x^2 + 24x + 64}}}.
Area of walkway = outer area - inner area 
= {{{2x^2 + 24x + 64 - 2x^2}}}
= {{{24x + 64}}}


Equation:  Area of walkway equals twice inner area (area of the bed):
{{{24x + 64= 2*(2x^2)}}} .
{{{24x + 64 = 4x^2}}}


Quadratic equations, so set equal to zero by subtracting 24x and 64 from both sides.
{{{0 = 4x^2 - 24x - 64 = 0}}}


Factor out the common factor of 4:
{{{0=4(x^2 - 6x - 16) }}}


Factor the trinomial:
{{{0 = 4(x-8)(x+2) }}}


Solutions are x = 8 and x = -2, but x represents the side of a rectangle, so it cannot be negative. 

 
Final answer: 
x = 8  width of bed. 
2x = 16 length of bed.


Check: 
Dimensions of the walkway
x+8 = 16
2x+8= 24
Area of bed = {{{8*16}}} = 128 sq. ft.
Total area = {{{16*24}}} = 384 sq. ft.
Area of walk = 384 - 128 = 256 sq. ft.


Area of walk = 2(area of bed)
256 = 2(128)


R^2 at SCC</PRE>