Question 222959
Let x = length and let y = width.


Your 2 equations are:


2x + 2y = 500
xy = 21800


Solvle for y in both equations to get:


y = 250-x
y = 21800/x


Graph these equations to get what is shown below:


{{{graph(300,300,-1000,1000,-1000,1000,250-x,21800/x)}}}


The graphs of these 2 equations do not intersect, therefore you do not have a solution that is common to both.


If you solve these two equations simultaneously by substitution, it will lead to an equation of x^2 - 250x + 21800 = 0.


Solve this using the quadratic equation and you will wind up with imaginary roots.


This means there is no solution for x that is real that will satisfy both equations, I believe.


The graph of the two equations bears this out.


I'm not 100% sure that I'm right but it sure looks like that.


I did a test, however, to see what happens if I do get an intersection on the graph.


I doubled the perimeter to 1000.


Your 2 equations would become:


2x + 2y = 1000
xy = 21800


Solvle for y in both equations to get:


y = 500-x
y = 21800/x


Graph these equations to get what is shown below:


{{{graph(300,300,-1000,1000,-1000,1000,500-x,21800/x)}}}


You do get an intersection now which says there is a solution to this revised equation.


Solving it algebraically using the quadratic formula does provide real roots to the equation this time.


We have a quadratic equation of x^2 - 500x + 21800 = 0 which yields roots of:


x = 48.25758998
and:
x = 451.74241


When x is 48.25758998, y = 451.74241


When x is 451.74241, y = 48.25758998


2 * x + 2 * y = 1000 which is the perimeter of the revised equation.


x*y = 48... * 451... = 21800 which is the area of the revised equation.



My original analysis is proved correct because when I do get an intersection ojn the graph, I have real solutions, which means that there is no solution to the equations you posed originally.  Those are:


2x + 2y = 500
x*y = 21800