1. **Stating the problem:** We are given the function $$f(x) = \frac{1}{6}x^3 + \frac{3}{2}ax^2 - 36a^2x + 10$$ where $$a \in \mathbb{R}^+$$ (positive real numbers). 2. **Goal:** Analyze the function for critical points, extrema, and behavior depending on the parameter $$a$$. 3. **Formula and rules:** To find critical points, we compute the first derivative $$f'(x)$$ and set it equal to zero: $$f'(x) = 0$$ 4. **Calculate the first derivative:** $$f'(x) = \frac{d}{dx} \left( \frac{1}{6}x^3 + \frac{3}{2}ax^2 - 36a^2x + 10 \right) = \frac{1}{6} \cdot 3x^2 + \frac{3}{2}a \cdot 2x - 36a^2 = \frac{1}{2}x^2 + 3ax - 36a^2$$ 5. **Set the derivative equal to zero to find critical points:** $$\frac{1}{2}x^2 + 3ax - 36a^2 = 0$$ Multiply both sides by 2 to clear the fraction: $$\cancel{2} \cdot \left( \frac{1}{2}x^2 + 3ax - 36a^2 \right) = \cancel{2} \cdot 0$$ $$x^2 + 6ax - 72a^2 = 0$$ 6. **Solve the quadratic equation:** Using the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a=1$$, $$b=6a$$, and $$c=-72a^2$$: $$x = \frac{-6a \pm \sqrt{(6a)^2 - 4 \cdot 1 \cdot (-72a^2)}}{2} = \frac{-6a \pm \sqrt{36a^2 + 288a^2}}{2} = \frac{-6a \pm \sqrt{324a^2}}{2}$$ 7. **Simplify the square root:** $$\sqrt{324a^2} = 18a$$ since $$a > 0$$. 8. **Find the two critical points:** $$x_1 = \frac{-6a + 18a}{2} = \frac{12a}{2} = 6a$$ $$x_2 = \frac{-6a - 18a}{2} = \frac{-24a}{2} = -12a$$ 9. **Determine the nature of critical points using the second derivative:** Calculate $$f''(x)$$: $$f''(x) = \frac{d}{dx} f'(x) = \frac{d}{dx} \left( \frac{1}{2}x^2 + 3ax - 36a^2 \right) = x + 3a$$ Evaluate at $$x_1 = 6a$$: $$f''(6a) = 6a + 3a = 9a > 0$$ so $$x=6a$$ is a local minimum. Evaluate at $$x_2 = -12a$$: $$f''(-12a) = -12a + 3a = -9a < 0$$ so $$x=-12a$$ is a local maximum. **Final answer:** The function $$f(x)$$ has a local maximum at $$x = -12a$$ and a local minimum at $$x = 6a$$ for $$a > 0$$.