Putnam

The answer is $\{a\,|\,a>2\}$. If $a>2$, then the function $f(x) = 2a/(a-2)$ has the desired property; both perimeter and area of $R$ in this case are $2a^2/(a-2)$. Now suppose that $a\leq 2$, and let $f(x)$ be a nonnegative continuous function on $[0,a]$. Let $P=(x_0,y_0)$ be a point on the graph of $f(x)$ with maximal $y$-coordinate; then the area of $R$ is at most $ay_0$ since it lies below the line $y=y_0$. On the other hand, the points $(0,0)$, $(a,0)$, and $P$ divide the boundary of $R$ into three sections. The length of the section between $(0,0)$ and $P$ is at least the distance between $(0,0)$ and $P$, which is at least $y_0$; the length of the section between $P$ and $(a,0)$ is similarly at least $y_0$; and the length of the section between $(0,0)$ and $(a,0)$ is $a$. Since $a\leq 2$, we have $2y_0 + a > ay_0$ and hence the perimeter of $R$ is strictly greater than the area of $R$.