jmc

Let the dimensions of the rectangle be $l$ and $w$. We are given $2l+2w=30$, which implies $l+w=15$. We want to maximize the product $lw$. We make this product maximal for a fixed sum when $l$ and $w$ are as close as possible. Since $l$ and $w$ are integers, they must be 7 and 8, which gives us a product of $\boxed{56}$.

Below is proof that we want $l$ and $w$ to be as close as possible.

Since $l+w=15$, we have $w=15-l$. The area of the rectangle is $lw=l(15-l)$. Completing the square gives $$\begin{array}{rl}& l(15-l)=15l-l^2=-(l^2-15l)\\ & =-\left(l^2-15l+{\left(\frac{15}{2}\right)}^2\right)+{\left(\frac{15}{2}\right)}^2\\ & =-{\left(l-\frac{15}{2}\right)}^2+{\left(\frac{15}{2}\right)}^2.\end{array}$$ Therefore, the area of the rectangle is $\frac{225}{4}$ minus the squared quantity ${\left(l-\frac{15}{2}\right)}^2$. So, we need $l$ to be as close to $\frac{15}{2}$ as possible to make this area as great as possible. Letting $l=7$ or $l=8$ gives us our maximum area, which is $8\cdot 7=\boxed{56}$.

Note that we might also have figured out the value of $l$ that gives us the maximum of $l(15-l)$ by considering the graph of $y=x(15-x)$. The graph of this equation is a parabola with $x$-intercepts $(0,0)$ and $(15,0)$. The axis of symmetry is mid-way between these intercepts, so it is at $x=7.5$, which means the vertex is on the line $x=7.5$. The parabola goes downward from the vertex both to the left and right, so the highest possible point on the graph that has an integer coordinate for $x$ must have $x=7$ or $x=8$ as the $x$-coordinate. So, the rectangle's length must be 7 or 8, as before.