Question 222959: how to figure out different lenghts and widths of a rectangle with perimeters of 500 ft and areas of 21,800 square feet
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 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:
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:
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
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