Question 24591
Is this for a calculus class?
You don't need calculus to solve it anyway
Always draw a picture

|------------------|
|wwwwwwwwarew|
|wwwwwwhouseww|  
|------------------|
|aaaafencedaaaaaa|   
|------------------|

call the sides of the fenced area x. both sides are equal in a rectangle
the sum of the 2 sides is 2x
if there are 120m of fencing, the third side is 120 - 2x
so the equation for area is
{{{A = x*(120 - 2*x)}}}
expanding
{{{A = 120 * x - 2*x^2}}}
try plugging in values for x to see what A is
x = 10
{{{A = 120 * 10 - 2 * (10^2)}}}
A = 1000
x = 20
{{{A = 120 * 20 - 2 * (20^2)}}}
A = 1600
x = 40
{{{A = 120 * 40 - 2 * (40^2)}}}
A = 1600
I get the same answer for x = 20 and x = 40, so the max must be in between,
since the curve is a parabola
x = 30
{{{A = 120 * 30 - 2 * (30^2)}}}
A = 1800
that is the max area in square meters, so the sides are 30 m and the third side
is 120 - 2(30) = 60 meters
check: 30 * 60 = 1800 m^2 for the area
and 2 * 30 + 60 = 120 for the length