1. **Stating the problem:** We have a rectangle where the length $L$ exceeds the width $W$ by 3 cm, i.e., $L = W + 3$. The area $A$ of the rectangle must not exceed 30 cm$^2$, so $A \leq 30$. 2. **Formula used:** The area of a rectangle is given by: $$A = L \times W$$ Since $L = W + 3$, substitute to get: $$A = W(W + 3)$$ 3. **Set up the inequality:** $$W(W + 3) \leq 30$$ which expands to: $$W^2 + 3W \leq 30$$ 4. **Rewrite the inequality:** $$W^2 + 3W - 30 \leq 0$$ 5. **Solve the quadratic inequality:** First, solve the quadratic equation: $$W^2 + 3W - 30 = 0$$ Using the quadratic formula: $$W = \frac{-3 \pm \sqrt{3^2 - 4 \times 1 \times (-30)}}{2 \times 1} = \frac{-3 \pm \sqrt{9 + 120}}{2} = \frac{-3 \pm \sqrt{129}}{2}$$ 6. **Calculate the roots:** $$\sqrt{129} \approx 11.36$$ So, $$W_1 = \frac{-3 + 11.36}{2} = \frac{8.36}{2} = 4.18$$ $$W_2 = \frac{-3 - 11.36}{2} = \frac{-14.36}{2} = -7.18$$ 7. **Determine the valid range for $W$:** Since width cannot be negative, discard $W_2 = -7.18$. The quadratic opens upwards (positive leading coefficient), so the inequality $W^2 + 3W - 30 \leq 0$ holds between the roots. Thus, $$-7.18 \leq W \leq 4.18$$ Only positive widths make sense, so: $$0 < W \leq 4.18$$ 8. **Find the corresponding length $L$:** $$L = W + 3$$ So, $$L \leq 4.18 + 3 = 7.18$$ **Final answer:** The width $W$ must be at most approximately 4.18 cm, and the length $L$ at most approximately 7.18 cm, to ensure the area does not exceed 30 cm$^2$.