1. **Problem Statement:** Correct and solve the first problem: Find the domain, global maximum, and global minimum of the function $$f(x,y) = \sqrt{64 - x^2 - y^2}$$. 2. **Domain of the function:** The function involves a square root, so the expression inside must be non-negative: $$64 - x^2 - y^2 \geq 0$$ This implies: $$x^2 + y^2 \leq 64$$ So, the domain is the disk centered at the origin with radius 8: $$D = \{(x,y) \mid x^2 + y^2 \leq 64\}$$ 3. **Finding global maximum and minimum:** - Since the square root function is increasing, the maximum value of $$f$$ occurs when $$64 - x^2 - y^2$$ is maximized. - The maximum of $$64 - x^2 - y^2$$ is 64, achieved at $$x=0, y=0$$. - So, the global maximum is: $$f(0,0) = \sqrt{64} = 8$$ - The minimum value occurs when $$64 - x^2 - y^2$$ is minimized within the domain, i.e., when $$x^2 + y^2 = 64$$. - At the boundary, $$f(x,y) = \sqrt{0} = 0$$. 4. **Summary:** - Domain: $$x^2 + y^2 \leq 64$$ - Global maximum: $$8$$ at $$(0,0)$$ - Global minimum: $$0$$ on the circle $$x^2 + y^2 = 64$$ 5. **Additional notes:** - The function is continuous and smooth inside the domain. - The domain is a closed and bounded set, so the global extrema exist.