Question 159212
it appears that the length can be any of 2 numbers and satisfy the equation.
the equation is
perimeter = l + 2*w = 18 (fence only covers 1 length)
area = l*w = 38
in the first equation, we take l = 18 - (2*w) and substitute for l in the second equation.
the second equation becomes
(18 - 2*w)*w = 38
this expands to
18*w - 2*w^2 = 38
subtracting both sides of the equation by 18*w and adding 2*w to both sides of the equation and it becomes
0 = 2*w^2 - 18*w + 38
which is the same as
{{{2*w^2 - 18*w + 38 = 0}}}
the answer is not apparent right away so we'll use the quadratic equation to solve.
quadratic equation is {{{w=((-b)+sqrt(b^2-4*a*c))/(2*a)}}} and {{{w=((-b)-sqrt(b^2-4*a*c))/(2*a)}}}.
(-b) = +18
a = 2
c = 38
{{{sqrt(b^2-4*a*c) = sqrt(20)}}}
{{{w = (18 + sqrt(20))/4)}}} = 5.618033989
{{{w = (18 - sqrt(20))/4)}}} = 3.381966011
solving for l, we get
18 - 2*5.618033989 = 6.763932923
18 - 2*3.381966011 = 11.23606798
width can be 5.618033989 while length is 6.763932923, or
width can be 3.381966011 while length is 11.23606798
to prove,
perimeter = l + 2*w = 18 (fence perimeter contained only 1*l)
6.763932923 + 2*5.618033989 = 18.0000009 = 18 if the numbers weren't rounded because of the limitation of the calculator display.
also
11.23606798 + 2*3.381966011 = 18
perimeter is ok.
area = 2*l + 2*w
6.763932923 * 5.618033989 = 37.99998431 = 38 if the numbers weren't rounded because of the limitation of the calculator display.
11.23606798 * 3.381966011 = 38.00000001 = 38 if the numbers weren't rounded because of the limitation of the calculator display.
to prove that for yourself, do the calculations using a calculator and store the results without truncating and then solve again using the numbers from inside the calculator rather than the numbers in the display.