Algebra 2 (1st Edition)

Our equation is according to the text of the exercise: $3\cdot18\cdot15=(18+x)(15+x)\\810=270+x^2+3x\\x^2+33x-540=0\\(x+45)(x-12)=0$ Thus $x=-45$ or $x=12$ but $x$ has to be positive, thus $x=12$. Thus the new dimensions are $30$ and $27$, and the length of rope needed is $2(30+27)=114ft$