A rectangle has a perimeter of 30 units and its dimensions are whole numbers. What is the maximum possible area of the rectangle in square units?
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{align*}
&l(15-l) = 15l-l^2 = -(l^2 - 15l) \\
&\qquad= -\left(l^2 - 15l +\left(\frac{15}{2}\right)^2\right) + \left(\frac{15}{2}\right)^2\\
&\qquad= -\left(l-\frac{15}{2}\right)^2 + \left(\frac{15}{2}\right)^2.\end{align*} 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.

[asy]
import graph; defaultpen(linewidth(0.8));
size(150,IgnoreAspect);
real f(real x)
{

return x*(15-x);
}
xaxis(Arrows(4));
yaxis(ymax=f(7.5),Arrows(4));
draw(graph(f,-3,18),Arrows(4));
label("Area",(0,f(7.5)),N);
label("$l$",(18,0),S);[/asy]