Answer
$5\ in, 3\ in$
Work Step by Step
1. Let the dimensions be $x,y$, based on the given conditions, we can get:
$\begin{cases}2(x+y)=16 \\ xy=15 \end{cases}$
2. The 1st equation gives $y=8-x$ use it in the 2nd equation to get $8x-x^2=15$ or $x^2-8x+15$ or $(x-3)(x-5)=0$, thus $x=3,5$, finally, use the 2nd equation to get $y=5, 3$
3. We have the solution(s) as: $5\ in, 3\ in$
