Area of a Rectangle by its perimeter

Question

If I have a rectangle, with its perimiter 100m how do I show that the area A is given by the quadratic function

A = 50x - $x^2$

Where x is the length of one side

Answer 1

Suppose $y$ denotes the length of the other side. Then, since the perimeter is $100$, we have

\begin{align} 2(x+y)&=100\\\Rightarrow y&=50-x. \end{align}

Thus, the area of the rectangle is

\begin{align} A&=xy\\&=x(50-x)\\&=50x-x^{2}. \end{align}

Answer 2

Consider a $x$ by $y$ rectangle. You've given its perimeter is $100$ and hence $x+x+y+y=2x+2y=100$. It's area is given by $A=xy$. Express $y$ in terms of $x$ to get rid of the $y$ in the equation of the area using the equation of the perimeter: $2x+2y=100$, and substitute it back in for y in $A=xy$.

Answer 3

Let $x,y$ be the side of your rectangle. You know that $2x + 2y = 100$. Also the area is $A = xy$

From the first equation you can say that $x = \frac{100-2y}{2} = 50 -y$, therefore $A = (50 -y)y$. Here the name $x$ and $y$ are arbitrary and all the equations are symmetric if you swap $x$ and $y$ so this works both for $x$ and for $y$.