## Precalculus (6th Edition) Blitzer

Consider the length and breadth of the rectangle to be $l$ and $b$ respectively. It is provided that the perimeter of the rectangle is $72$ feet. Consider the perimeter to be denoted by $p$, so: $p=2\left( l+b \right)$ Substitute the value of $p$ in the above equation: \begin{align} & 72=2\left( l+b \right) \\ & l+b=36 \end{align} Also, it is provided that the length of the rectangle is 14 feet more than the width: $l=b+14$ So, \begin{align} & l+b=36 \\ & b+14+b=36 \\ & 2b=36-14 \\ & b=11 \end{align} So, the length will be: \begin{align} & l=b+14 \\ & =11+14 \\ & =25 \end{align} The dimensions of the rectangles are 25 feet by 11 feet.