Question 35452
Let x = width of the rectangle
y = length of the rectangle


Two equations are given in this problem:
Area = xy = 3600 square feet
Perimeter = 2x+2y = 300 feet


In the second equation, it will be easy to solve for y by dividing both sides by 2:
x+y = 150

y= 150-x


Substitute this back into the first equation:

{{{xy = 3600}}}
{{{x(150-x) = 3600}}}
{{{150x - x^2 = 3600}}}


This is a quadratic equation.  Set the equation equal to zero, by adding {{{+x^2 -150x}}} to each side of the equation.
{{{0 = x^2 - 150x +3600}}}


Does it factor???  Probably so!
{{{0=(x-30)(x-90)}}}
x=30 or x= 120


If x = 30, then y = 120, and if x= 120, then y = 30.  It would be appropriate to say that the width would be the smaller number x= 30 feet, and the length is the larger number, which would be y = 120 feet.


R^2 at SCC