1. **Problem statement:** Find the stationary points and their nature for the functions: (a) $$z = 2x^2y^2 + 4xy^2 - 4y^3 + 16y + 5$$ (b) $$z = 4 - 25x^2 + 20xy - 4y^2$$ 2. **Stationary points** occur where the first partial derivatives with respect to $x$ and $y$ are zero: $$\frac{\partial z}{\partial x} = 0, \quad \frac{\partial z}{\partial y} = 0$$ 3. **(a) For** $$z = 2x^2y^2 + 4xy^2 - 4y^3 + 16y + 5$$ Calculate partial derivatives: $$\frac{\partial z}{\partial x} = 4xy^2 + 4y^2 = 4y^2(x + 1)$$ $$\frac{\partial z}{\partial y} = 4x^2y + 8xy - 12y^2 + 16$$ Set derivatives to zero: $$4y^2(x + 1) = 0 \implies y=0 \text{ or } x=-1$$ For $y=0$: $$\frac{\partial z}{\partial y} = 16 \neq 0$$ No stationary point here. For $x=-1$: $$4(-1)^2 y + 8(-1) y - 12 y^2 + 16 = 4y - 8y - 12 y^2 + 16 = -4y - 12 y^2 + 16 = 0$$ Divide by $-4$: $$\cancel{-4}y + \cancel{-12} y^2 - \cancel{16} = 0 \implies y + 3 y^2 - 4 = 0$$ Rearranged: $$3 y^2 + y - 4 = 0$$ Solve quadratic: $$y = \frac{-1 \pm \sqrt{1 + 48}}{6} = \frac{-1 \pm 7}{6}$$ Two solutions: $$y = 1 \quad \text{or} \quad y = -\frac{4}{3}$$ Stationary points: $$(x,y) = (-1,1), (-1,-\frac{4}{3})$$ 4. **Determine nature using second derivatives:** $$z_{xx} = \frac{\partial^2 z}{\partial x^2} = 4 y^2$$ $$z_{yy} = \frac{\partial^2 z}{\partial y^2} = 4 x^2 + 8 x - 24 y$$ $$z_{xy} = \frac{\partial^2 z}{\partial x \partial y} = 8 x y + 8 y$$ Calculate discriminant: $$D = z_{xx} z_{yy} - (z_{xy})^2$$ At $(-1,1)$: $$z_{xx} = 4(1)^2 = 4$$ $$z_{yy} = 4(1) + 8(-1) - 24(1) = 4 - 8 - 24 = -28$$ $$z_{xy} = 8(-1)(1) + 8(1) = -8 + 8 = 0$$ $$D = 4 \times (-28) - 0^2 = -112 < 0$$ Saddle point at $(-1,1)$. At $(-1,-\frac{4}{3})$: $$z_{xx} = 4 \left(-\frac{4}{3}\right)^2 = 4 \times \frac{16}{9} = \frac{64}{9}$$ $$z_{yy} = 4(1) + 8(-1) - 24 \left(-\frac{4}{3}\right) = 4 - 8 + 32 = 28$$ $$z_{xy} = 8(-1) \left(-\frac{4}{3}\right) + 8 \left(-\frac{4}{3}\right) = \frac{32}{3} - \frac{32}{3} = 0$$ $$D = \frac{64}{9} \times 28 - 0 = \frac{1792}{9} > 0$$ Since $z_{xx} > 0$ and $D > 0$, this is a local minimum. --- 5. **(b) For** $$z = 4 - 25x^2 + 20xy - 4y^2$$ Calculate partial derivatives: $$\frac{\partial z}{\partial x} = -50 x + 20 y$$ $$\frac{\partial z}{\partial y} = 20 x - 8 y$$ Set to zero: $$-50 x + 20 y = 0 \implies 20 y = 50 x \implies y = \frac{50}{20} x = \frac{5}{2} x$$ $$20 x - 8 y = 0$$ Substitute $y$: $$20 x - 8 \times \frac{5}{2} x = 20 x - 20 x = 0$$ This holds for all $x$, so stationary points lie on the line: $$y = \frac{5}{2} x$$ 6. **Second derivatives:** $$z_{xx} = -50, \quad z_{yy} = -8, \quad z_{xy} = 20$$ Discriminant: $$D = z_{xx} z_{yy} - (z_{xy})^2 = (-50)(-8) - 20^2 = 400 - 400 = 0$$ Since $D=0$, the test is inconclusive; the stationary points form a line of saddle points or a degenerate case. --- **Final answers:** (a) Stationary points at $(-1,1)$ (saddle point) and $\left(-1,-\frac{4}{3}\right)$ (local minimum). (b) Stationary points lie on the line $y = \frac{5}{2} x$; nature is degenerate with $D=0$.