1. **Problem statement:** A rectangle has a perimeter of 74. The length ($l$) and width ($w$) are integers. We need to find the maximum possible area of this rectangle. 2. **Formula for perimeter:** The perimeter $P$ of a rectangle is given by $$P = 2(l + w)$$ Given $P = 74$, we have $$2(l + w) = 74$$ which simplifies to $$l + w = 37$$ 3. **Formula for area:** The area $A$ of a rectangle is $$A = l \times w$$ 4. **Express area in terms of one variable:** Since $w = 37 - l$, substitute into area formula: $$A = l(37 - l) = 37l - l^2$$ 5. **Maximize the area:** The area is a quadratic function of $l$: $$A(l) = -l^2 + 37l$$ This is a downward opening parabola, so the maximum occurs at the vertex. 6. **Find vertex:** The vertex $l$-value is given by $$l = -\frac{b}{2a} = -\frac{37}{2(-1)} = \frac{37}{2} = 18.5$$ 7. **Since $l$ and $w$ must be integers, check integers near 18.5:** - For $l=18$, $w=37-18=19$, area $= 18 \times 19 = 342$ - For $l=19$, $w=37-19=18$, area $= 19 \times 18 = 342$ 8. **Conclusion:** The maximum possible area with integer sides is $$\boxed{342}$$