1. The problem is to analyze the function $$y = \frac{1}{2}x^2 \cdot \sqrt{16 - x^2}$$. 2. First, understand the domain: since there is a square root $$\sqrt{16 - x^2}$$, we require $$16 - x^2 \geq 0$$, which implies $$-4 \leq x \leq 4$$. 3. The function involves $$\frac{1}{2} x^2$$ multiplied by $$\sqrt{16 - x^2}$$. 4. To simplify or study critical points, consider the derivative using the product and chain rules. 5. Let $$f(x) = \frac{1}{2} x^2$$ and $$g(x) = \sqrt{16 - x^2} = (16 - x^2)^{\frac{1}{2}}$$. 6. Then, $$f'(x) = x$$. 7. For $$g(x)$$, $$g'(x) = \frac{1}{2} (16 - x^2)^{-\frac{1}{2}} \cdot (-2x) = -\frac{x}{\sqrt{16 - x^2}}$$. 8. By the product rule: $$y' = f'(x) g(x) + f(x) g'(x) = x \sqrt{16 - x^2} + \frac{1}{2} x^2 \left(-\frac{x}{\sqrt{16 - x^2}}\right) = x \sqrt{16 - x^2} - \frac{1}{2} \frac{x^3}{\sqrt{16 - x^2}}$$. 9. To find critical points, set $$y' = 0$$: $$x \sqrt{16 - x^2} - \frac{1}{2} \frac{x^3}{\sqrt{16 - x^2}} = 0$$. 10. Multiply both sides by $$\sqrt{16 - x^2}$$ to clear denominators: $$x (16 - x^2) - \frac{1}{2} x^3 = 0$$. 11. Simplify: $$16x - x^3 - \frac{1}{2} x^3 = 0 \Rightarrow 16x - \frac{3}{2} x^3 = 0$$. 12. Factor out $$x$$: $$x \left(16 - \frac{3}{2} x^2\right) = 0$$. 13. So critical points at $$x = 0$$ or when $$16 - \frac{3}{2} x^2 = 0$$. 14. Solve for $$x$$: $$16 = \frac{3}{2} x^2 \Rightarrow x^2 = \frac{16 \times 2}{3} = \frac{32}{3}$$. 15. Thus, $$x = \pm \sqrt{\frac{32}{3}} = \pm \frac{4\sqrt{6}}{3} \approx \pm 3.266$$. 16. All critical points lie within the domain $$[-4,4]$$. 17. Evaluate $$y$$ at critical points: - At $$x=0$$, $$y = \frac{1}{2} \cdot 0^2 \cdot \sqrt{16-0} = 0$$. - At $$x= \frac{4\sqrt{6}}{3}$$, $$y = \frac{1}{2} x^2 \sqrt{16 - x^2} = \frac{1}{2} \cdot \frac{32}{3} \cdot \sqrt{16 - \frac{32}{3}} = \frac{16}{3} \cdot \sqrt{\frac{16}{3}} = \frac{16}{3} \cdot \frac{4}{\sqrt{3}} = \frac{64}{3 \sqrt{3}} = \frac{64 \sqrt{3}}{9} \approx 12.31$$. - Similarly, for $$x = -\frac{4\sqrt{6}}{3}$$, $$y$$ is the same due to squaring. 18. Summary: the function has zeros at $$x=\pm 4$$ and $$x=0$$, peaks at approximately $$x=\pm 3.266$$ with max value about 12.31. Final Answer: The maximum value of $$y$$ in the domain $$[-4,4]$$ is approximately $$12.31$$ at $$x=\pm \frac{4\sqrt{6}}{3}$$.