1. Problem: Find all local maxima and minima of $$f(x,y) = x^2 + 4y^2 - 2x + 8y - 1$$. Step 1: Find partial derivatives: $$f_x = 2x - 2, \quad f_y = 8y + 8$$ Step 2: Set derivatives to zero to find critical points: $$2x - 2 = 0 \Rightarrow x = 1$$ $$8y + 8 = 0 \Rightarrow y = -1$$ Step 3: Compute second derivatives: $$f_{xx} = 2, \quad f_{yy} = 8, \quad f_{xy} = 0$$ Step 4: Compute discriminant: $$D = f_{xx} f_{yy} - (f_{xy})^2 = 2 \times 8 - 0 = 16 > 0$$ Since $$f_{xx} > 0$$ and $$D > 0$$, the point $$(1,-1)$$ is a local minimum. 2. Problem: Find all local maxima and minima of $$f(x,y) = x^2 y - y^3 + 6x - 10y + 2$$. Step 1: Find partial derivatives: $$f_x = 2xy + 6, \quad f_y = x^2 - 3y^2 - 10$$ Step 2: Set derivatives to zero: $$2xy + 6 = 0 \Rightarrow 2xy = -6 \Rightarrow xy = -3$$ $$x^2 - 3y^2 - 10 = 0$$ Step 3: From $$xy = -3$$, express $$x = -3/y$$ (assuming $$y \neq 0$$). Step 4: Substitute into second equation: $$\left(-\frac{3}{y}\right)^2 - 3y^2 - 10 = 0 \Rightarrow \frac{9}{y^2} - 3y^2 - 10 = 0$$ Multiply both sides by $$y^2$$: $$9 - 3y^4 - 10y^2 = 0 \Rightarrow 3y^4 + 10y^2 - 9 = 0$$ Step 5: Let $$z = y^2$$, then: $$3z^2 + 10z - 9 = 0$$ Step 6: Solve quadratic: $$z = \frac{-10 \pm \sqrt{100 + 108}}{6} = \frac{-10 \pm \sqrt{208}}{6}$$ $$\sqrt{208} \approx 14.42$$ Positive root: $$z = \frac{-10 + 14.42}{6} = \frac{4.42}{6} \approx 0.737$$ Negative root is discarded since $$z = y^2 \geq 0$$. Step 7: So $$y = \pm \sqrt{0.737} \approx \pm 0.859$$. Step 8: Find corresponding $$x$$: $$x = -3/y$$ For $$y = 0.859$$, $$x \approx -3/0.859 = -3.49$$ For $$y = -0.859$$, $$x \approx -3/(-0.859) = 3.49$$ Step 9: Compute second derivatives: $$f_{xx} = 2y, \quad f_{yy} = -6y, \quad f_{xy} = 2x$$ Step 10: Evaluate discriminant: $$D = f_{xx} f_{yy} - (f_{xy})^2 = (2y)(-6y) - (2x)^2 = -12y^2 - 4x^2$$ At $$(x,y) = (-3.49, 0.859)$$: $$D = -12(0.859)^2 - 4(-3.49)^2 = -12(0.737) - 4(12.18) = -8.84 - 48.72 = -57.56 < 0$$ Since $$D < 0$$, this point is a saddle point. Similarly for $$(3.49, -0.859)$$, $$D < 0$$, also saddle. No local maxima or minima. 3. Problem: Find all local maxima and minima of $$f = 9 + 4x - y - 2x^2 - 3y^2$$. Step 1: Partial derivatives: $$f_x = 4 - 4x, \quad f_y = -1 - 6y$$ Step 2: Set to zero: $$4 - 4x = 0 \Rightarrow x = 1$$ $$-1 - 6y = 0 \Rightarrow y = -\frac{1}{6}$$ Step 3: Second derivatives: $$f_{xx} = -4, \quad f_{yy} = -6, \quad f_{xy} = 0$$ Step 4: Discriminant: $$D = (-4)(-6) - 0 = 24 > 0$$ Since $$f_{xx} < 0$$ and $$D > 0$$, point $$(1, -\frac{1}{6})$$ is a local maximum. 4. Problem: Find all local maxima and minima of $$f = x^2 + 4xy + y^2 - 6y + 1$$. Step 1: Partial derivatives: $$f_x = 2x + 4y, \quad f_y = 4x + 2y - 6$$ Step 2: Set to zero: $$2x + 4y = 0 \Rightarrow x = -2y$$ Substitute into second: $$4(-2y) + 2y - 6 = 0 \Rightarrow -8y + 2y - 6 = 0 \Rightarrow -6y = 6 \Rightarrow y = -1$$ Then $$x = -2(-1) = 2$$ Step 3: Second derivatives: $$f_{xx} = 2, \quad f_{yy} = 2, \quad f_{xy} = 4$$ Step 4: Discriminant: $$D = 2 \times 2 - 4^2 = 4 - 16 = -12 < 0$$ Since $$D < 0$$, the critical point $$(2, -1)$$ is a saddle point. 5. Problem: Find all local maxima and minima of $$f = x^3 + 3y^3 - 15x - 27y$$. Step 1: Partial derivatives: $$f_x = 3x^2 - 15, \quad f_y = 9y^2 - 27$$ Step 2: Set to zero: $$3x^2 - 15 = 0 \Rightarrow x^2 = 5 \Rightarrow x = \pm \sqrt{5}$$ $$9y^2 - 27 = 0 \Rightarrow y^2 = 3 \Rightarrow y = \pm \sqrt{3}$$ Step 3: Second derivatives: $$f_{xx} = 6x, \quad f_{yy} = 18y, \quad f_{xy} = 0$$ Step 4: Discriminant: $$D = f_{xx} f_{yy} - (f_{xy})^2 = 6x \times 18y - 0 = 108xy$$ Evaluate at each critical point: - At $$(\sqrt{5}, \sqrt{3})$$: $$D = 108 \times \sqrt{5} \times \sqrt{3} > 0$$ and $$f_{xx} = 6\sqrt{5} > 0$$, local minimum. - At $$(\sqrt{5}, -\sqrt{3})$$: $$D = 108 \times \sqrt{5} \times (-\sqrt{3}) < 0$$, saddle. - At $$(-\sqrt{5}, \sqrt{3})$$: $$D < 0$$, saddle. - At $$(-\sqrt{5}, -\sqrt{3})$$: $$D > 0$$ and $$f_{xx} = 6(-\sqrt{5}) < 0$$, local maximum. 6. Problem: Find all local maxima and minima of $$f = x^2 - xy + 2y^2 - 5x + 6y - 9$$. Step 1: Partial derivatives: $$f_x = 2x - y - 5, \quad f_y = -x + 4y + 6$$ Step 2: Set to zero: $$2x - y - 5 = 0 \Rightarrow y = 2x - 5$$ Substitute into second: $$-x + 4(2x - 5) + 6 = 0 \Rightarrow -x + 8x - 20 + 6 = 0 \Rightarrow 7x - 14 = 0 \Rightarrow x = 2$$ Then $$y = 2(2) - 5 = 4 - 5 = -1$$ Step 3: Second derivatives: $$f_{xx} = 2, \quad f_{yy} = 4, \quad f_{xy} = -1$$ Step 4: Discriminant: $$D = 2 \times 4 - (-1)^2 = 8 - 1 = 7 > 0$$ Since $$f_{xx} > 0$$ and $$D > 0$$, point $$(2, -1)$$ is a local minimum. 7. Problem: Find absolute max and min of $$f = x^2 + 3y - 3xy$$ over region bounded by $$y=0$$ and $$y=2$$. Step 1: Since $$y$$ is bounded, check critical points inside and on boundaries. Step 2: Partial derivatives: $$f_x = 2x - 3y, \quad f_y = 3 - 3x$$ Step 3: Set to zero: $$2x - 3y = 0 \Rightarrow y = \frac{2x}{3}$$ $$3 - 3x = 0 \Rightarrow x = 1$$ Step 4: Substitute $$x=1$$ into $$y=\frac{2x}{3}$$: $$y = \frac{2}{3}$$ which is in the region $$0 \leq y \leq 2$$. Step 5: Evaluate $$f$$ at critical point: $$f(1, \frac{2}{3}) = 1^2 + 3 \times \frac{2}{3} - 3 \times 1 \times \frac{2}{3} = 1 + 2 - 2 = 1$$ Step 6: Evaluate $$f$$ on boundaries: - At $$y=0$$: $$f = x^2 + 0 - 0 = x^2$$, minimum at $$x=0$$ with $$f=0$$, no max since $$x$$ unbounded. - At $$y=2$$: $$f = x^2 + 6 - 6x = x^2 - 6x + 6$$ Complete the square: $$= (x - 3)^2 - 9 + 6 = (x - 3)^2 - 3$$ Minimum at $$x=3$$ with $$f = -3$$, no max since $$x$$ unbounded. Step 7: Since $$x$$ is unbounded, no absolute max or min over entire region unless $$x$$ is bounded. 8. Problem: Minimize surface area of a box holding volume $$\frac{1}{2}$$ cubic meters. Step 1: Let dimensions be $$x,y,z$$. Volume constraint: $$xyz = \frac{1}{2}$$ Surface area: $$S = 2(xy + yz + zx)$$ Step 2: Use Lagrange multipliers or symmetry. Step 3: For minimum surface area with fixed volume, cube shape is optimal: $$x = y = z$$ Step 4: Then: $$x^3 = \frac{1}{2} \Rightarrow x = \sqrt[3]{\frac{1}{2}} = \frac{1}{\sqrt[3]{2}}$$ Step 5: So box should be a cube with side $$\frac{1}{\sqrt[3]{2}}$$ meters. 9. Problem: Maximize volume of rectangular box with length $$l$$ and girth $$g$$ such that $$l + g \leq 130$$ inches. Step 1: Let dimensions be $$l, w, h$$ with $$l$$ largest. Girth: $$g = 2(w + h)$$ Constraint: $$l + 2(w + h) = 130$$ Step 2: Volume: $$V = lwh$$ Step 3: Express $$l$$: $$l = 130 - 2(w + h)$$ Step 4: Volume: $$V = (130 - 2(w + h))wh = 130wh - 2w^2 h - 2w h^2$$ Step 5: Use symmetry to maximize volume, set $$w = h$$. Then: $$V = (130 - 4w) w^2 = 130 w^2 - 4 w^3$$ Step 6: Differentiate w.r.t $$w$$: $$\frac{dV}{dw} = 260 w - 12 w^2 = 0$$ Step 7: Solve: $$w(260 - 12 w) = 0 \Rightarrow w = 0 \text{ or } w = \frac{260}{12} = \frac{65}{3} \approx 21.67$$ Step 8: Compute $$l$$: $$l = 130 - 4w = 130 - 4 \times 21.67 = 130 - 86.68 = 43.32$$ Step 9: Volume: $$V = 43.32 \times 21.67 \times 21.67 \approx 20350$$ cubic inches. Step 10: Check second derivative to confirm max. Final answers summarized: 1. Local minimum at $$(1,-1)$$. 2. No local max/min, saddle points at approximately $$(\pm 3.49, \mp 0.859)$$. 3. Local maximum at $$(1, -\frac{1}{6})$$. 4. Saddle point at $$(2,-1)$$. 5. Local minimum at $$(\sqrt{5}, \sqrt{3})$$, local maximum at $$(-\sqrt{5}, -\sqrt{3})$$, others saddle. 6. Local minimum at $$(2,-1)$$. 7. No absolute max/min without $$x$$ bounds; critical point at $$(1, \frac{2}{3})$$. 8. Cube with side $$\frac{1}{\sqrt[3]{2}}$$ meters minimizes surface area. 9. Largest volume approx 20350 cubic inches with $$w = h \approx 21.67$$ inches, $$l \approx 43.32$$ inches.