1. **State the problem:** We have a concave mirror with radius of curvature $R = 36$ cm. An erect image is formed that is three times the size of the object. We need to find the position of the object. 2. **Recall mirror formulas:** The magnification $m$ is given by $$m = -\frac{v}{u}$$ where $v$ is the image distance and $u$ is the object distance. Since the image is erect and magnified three times, $m = +3$ (positive because erect). 3. **Find focal length:** The focal length $f$ is related to radius of curvature by $$f = \frac{R}{2} = \frac{36}{2} = 18 \text{ cm}$$ 4. **Use mirror equation:** $$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$ 5. **Express $v$ in terms of $u$ using magnification:** $$m = -\frac{v}{u} \Rightarrow v = -m u = -3 u$$ 6. **Substitute $v$ into mirror equation:** $$\frac{1}{18} = \frac{1}{u} + \frac{1}{-3u} = \frac{1}{u} - \frac{1}{3u} = \frac{2}{3u}$$ 7. **Solve for $u$:** $$\frac{1}{18} = \frac{2}{3u} \Rightarrow 3u = 2 \times 18 \Rightarrow 3u = 36 \Rightarrow u = 12 \text{ cm}$$ 8. **Interpretation:** The object is placed 12 cm in front of the mirror (since $u$ is positive for real objects in mirror conventions). **Final answer:** The object is located at $12$ cm from the mirror.