Question 574350
With the choices, this is not a very difficult problem.
If the fenced area has to be a rectangle, we want the perimeter to be 24 feet because to get the largest fenced area we want to use all the fencing available.
Half of the perimeter (12 feet) would be the sum of the lengths of two adjacent sides (maybe a long side plus a short side).
For a rectangle 12 feet long by 4 feet wide we would need
2(12+4) = 32 feet of fencing, so the last option does not work, because we do not have enough fencing.
The first three choices could be used. We have enough fencing, because
9+3=12
10+2=12 and
6+6=12
The question is which would give you the largest area.
Area is calculated as length times width, so the areas of those three rectangles, in square feet would be
{{{9*3=27}}}
{{{10*2=20}}} and
{{{6*6=36}}}
So, a square, 6 feet by 6 feet, is the best of the choices given, the one with the greatest area.
Without the choices, this is a quadratic function/parabola problem, and that's pretty advanced algebra.
Assuming that it has to be a rectangle, we would call the lengths of the sides {{{x}}} and {{{y}}}, and would say that
{{{x+y=12}}} --> {{{y=12-x}}}, and the area would be
{{{Area=x*y=x(12-x)=-x^2+12x}}}
So the area ys a quadratic function of x, which would graph as a parabola.
The maximum is at
{{{x=-12/(2*(-1))=6}}}, so {{{highlight(x=6)}}}. Then, {{{highlight(y=6)}}} too, because {{{y=12-6=6}}}