1. **Problem 1:** Examine the function $$z = 8x^3 + 2y - 3x^2 + y^2 + 1$$ for maximum, minimum, or saddle points. 2. **Step 1: Find the first partial derivatives.** $$f_x = \frac{\partial z}{\partial x} = 24x^2 - 6x$$ $$f_y = \frac{\partial z}{\partial y} = 2 + 2y$$ 3. **Step 2: Set the first derivatives equal to zero to find critical points.** $$24x^2 - 6x = 0 \implies 6x(4x - 1) = 0 \implies x = 0 \text{ or } x = \frac{1}{4}$$ $$2 + 2y = 0 \implies y = -1$$ 4. **Step 3: Critical points are at** $$(0, -1) \text{ and } \left(\frac{1}{4}, -1\right)$$ 5. **Step 4: Find the second partial derivatives.** $$f_{xx} = 48x - 6$$ $$f_{yy} = 2$$ $$f_{xy} = 0$$ 6. **Step 5: Use the second derivative test.** Calculate the discriminant $$D = f_{xx} f_{yy} - (f_{xy})^2 = (48x - 6)(2) - 0 = 2(48x - 6)$$ - At $x=0$: $$D = 2(0 - 6) = -12 < 0$$, so saddle point at $(0, -1)$. - At $x=\frac{1}{4}$: $$D = 2(48 \times \frac{1}{4} - 6) = 2(12 - 6) = 12 > 0$$ and $$f_{xx} = 48 \times \frac{1}{4} - 6 = 12 - 6 = 6 > 0$$, so local minimum at $\left(\frac{1}{4}, -1\right)$. 7. **Step 6: Find the function values at critical points.** $$z(0, -1) = 8(0)^3 + 2(-1) - 3(0)^2 + (-1)^2 + 1 = 0 - 2 + 0 + 1 + 1 = 0$$ $$z\left(\frac{1}{4}, -1\right) = 8\left(\frac{1}{4}\right)^3 + 2(-1) - 3\left(\frac{1}{4}\right)^2 + (-1)^2 + 1 = 8\times \frac{1}{64} - 2 - 3\times \frac{1}{16} + 1 + 1 = \frac{1}{8} - 2 - \frac{3}{16} + 2 = \frac{1}{8} - \frac{3}{16} = \frac{2}{16} - \frac{3}{16} = -\frac{1}{16}$$ --- 8. **Problem 2:** Examine the function $$z = 4x^2 - 3y + y^2 - x^2 = 3x^2 - 3y + y^2$$ for maximum, minimum, or saddle points. 9. **Step 1: Find the first partial derivatives.** $$f_x = 6x$$ $$f_y = -3 + 2y$$ 10. **Step 2: Set the first derivatives equal to zero.** $$6x = 0 \implies x = 0$$ $$-3 + 2y = 0 \implies y = \frac{3}{2}$$ 11. **Step 3: Critical point at** $$(0, \frac{3}{2})$$ 12. **Step 4: Find the second partial derivatives.** $$f_{xx} = 6$$ $$f_{yy} = 2$$ $$f_{xy} = 0$$ 13. **Step 5: Calculate the discriminant.** $$D = f_{xx} f_{yy} - (f_{xy})^2 = 6 \times 2 - 0 = 12 > 0$$ Since $$f_{xx} = 6 > 0$$, the critical point is a local minimum. 14. **Step 6: Find the function value at the critical point.** $$z\left(0, \frac{3}{2}\right) = 3(0)^2 - 3\times \frac{3}{2} + \left(\frac{3}{2}\right)^2 = 0 - \frac{9}{2} + \frac{9}{4} = -\frac{9}{2} + \frac{9}{4} = -\frac{9}{4} = -2.25$$ **Final answers:** - For problem 1, saddle point at $(0, -1)$ with $z=0$ and local minimum at $\left(\frac{1}{4}, -1\right)$ with $z = -\frac{1}{16}$. - For problem 2, local minimum at $(0, \frac{3}{2})$ with $z = -\frac{9}{4}$.