1. **Problem 1:** Find the minimum and maximum values of $f(x,y) = xy$ subject to the constraint $2x^2 + 8y^2 = 16$. 2. **Method:** Use Lagrange multipliers. Set up the system: $$\nabla f = \lambda \nabla g$$ where $g(x,y) = 2x^2 + 8y^2 - 16 = 0$. 3. Compute gradients: $$\nabla f = \left( y, x \right), \quad \nabla g = \left( 4x, 16y \right)$$ 4. Set equations: $$y = \lambda 4x, \quad x = \lambda 16y$$ 5. From the first, $\lambda = \frac{y}{4x}$ (assuming $x \neq 0$). From the second, $\lambda = \frac{x}{16y}$ (assuming $y \neq 0$). 6. Equate: $$\frac{y}{4x} = \frac{x}{16y} \implies 16y^2 = 4x^2 \implies 4y^2 = x^2$$ 7. Substitute into constraint: $$2x^2 + 8y^2 = 16 \implies 2x^2 + 8 \cdot \frac{x^2}{4} = 16 \implies 2x^2 + 2x^2 = 16 \implies 4x^2 = 16 \implies x^2 = 4$$ 8. Then $y^2 = \frac{x^2}{4} = 1$. 9. Points: $(x,y) = (\pm 2, \pm 1)$. 10. Evaluate $f$ at these points: $$f(2,1) = 2 \times 1 = 2, \quad f(2,-1) = -2, \quad f(-2,1) = -2, \quad f(-2,-1) = 2$$ 11. So maximum value is 2, minimum is -2. --- 12. **Problem 2:** $f(x,y) = x^2 - y^2$, constraint $x^2 + y^2 = 25$. 13. Gradients: $$\nabla f = (2x, -2y), \quad \nabla g = (2x, 2y)$$ 14. Set: $$2x = \lambda 2x, \quad -2y = \lambda 2y$$ 15. For $x \neq 0$, $2x = 2\lambda x \implies \lambda = 1$. 16. For $y \neq 0$, $-2y = 2\lambda y \implies \lambda = -1$. 17. Contradiction unless $x=0$ or $y=0$. 18. Case 1: $x=0$, then $y^2=25$, points $(0, \pm 5)$, $f = 0 - 25 = -25$. 19. Case 2: $y=0$, then $x^2=25$, points $(\pm 5, 0)$, $f = 25 - 0 = 25$. 20. So max is 25 at $(\pm 5,0)$, min is -25 at $(0, \pm 5)$. --- 21. **Problem 3:** $f=4x^3 + y^2$, constraint $2x^2 + y^2 = 1$. 22. Gradients: $$\nabla f = (12x^2, 2y), \quad \nabla g = (4x, 2y)$$ 23. Set: $$12x^2 = \lambda 4x, \quad 2y = \lambda 2y$$ 24. For $x \neq 0$, $12x^2 = 4\lambda x \implies 3x = \lambda$. 25. For $y \neq 0$, $2y = 2\lambda y \implies 1 = \lambda$. 26. Equate $\lambda$: $3x = 1 \implies x = \frac{1}{3}$. 27. Substitute into constraint: $$2 \left( \frac{1}{3} \right)^2 + y^2 = 1 \implies \frac{2}{9} + y^2 = 1 \implies y^2 = \frac{7}{9}$$ 28. Points: $(\frac{1}{3}, \pm \frac{\sqrt{7}}{3})$. 29. Evaluate $f$: $$f = 4 \left( \frac{1}{3} \right)^3 + y^2 = 4 \cdot \frac{1}{27} + \frac{7}{9} = \frac{4}{27} + \frac{21}{27} = \frac{25}{27}$$ 30. Check $y=0$ case: if $y=0$, then $2x^2=1 \implies x=\pm \frac{1}{\sqrt{2}}$, $f=4x^3 = 4 \left( \pm \frac{1}{\sqrt{2}} \right)^3 = \pm \frac{4}{2\sqrt{2}} = \pm \frac{2\sqrt{2}}{4} = \pm \frac{\sqrt{2}}{2}$ approx $\pm 0.707$. 31. So max approx $0.9259$ at $(\frac{1}{3}, \pm \frac{\sqrt{7}}{3})$, min approx $-0.707$ at $(-\frac{1}{\sqrt{2}},0)$. --- 32. **Problem 4:** $f = x - 3y - 1$, constraint $x^2 + 3y^2 = 16$. 33. Gradients: $$\nabla f = (1, -3), \quad \nabla g = (2x, 6y)$$ 34. Set: $$1 = \lambda 2x, \quad -3 = \lambda 6y$$ 35. From first: $\lambda = \frac{1}{2x}$, second: $\lambda = \frac{-3}{6y} = \frac{-1}{2y}$. 36. Equate: $$\frac{1}{2x} = \frac{-1}{2y} \implies y = -x$$ 37. Substitute into constraint: $$x^2 + 3(-x)^2 = 16 \implies x^2 + 3x^2 = 16 \implies 4x^2 = 16 \implies x^2 = 4$$ 38. So $x = \pm 2$, $y = \mp 2$. 39. Evaluate $f$: $$f(2,-2) = 2 - 3(-2) - 1 = 2 + 6 - 1 = 7$$ $$f(-2,2) = -2 - 6 - 1 = -9$$ 40. Max is 7 at $(2,-2)$, min is -9 at $(-2,2)$. --- 41. **Problem 5:** $f = 2x + y - 2z$, constraint $x^2 + y^2 + z^2 = 4$. 42. Gradients: $$\nabla f = (2,1,-2), \quad \nabla g = (2x, 2y, 2z)$$ 43. Set: $$2 = \lambda 2x, \quad 1 = \lambda 2y, \quad -2 = \lambda 2z$$ 44. Solve for $x,y,z$: $$x = \frac{1}{\lambda}, \quad y = \frac{1}{2\lambda}, \quad z = -\frac{1}{\lambda}$$ 45. Substitute into constraint: $$\left( \frac{1}{\lambda} \right)^2 + \left( \frac{1}{2\lambda} \right)^2 + \left( -\frac{1}{\lambda} \right)^2 = 4$$ $$\frac{1}{\lambda^2} + \frac{1}{4\lambda^2} + \frac{1}{\lambda^2} = 4$$ $$\frac{1 + \frac{1}{4} + 1}{\lambda^2} = 4 \implies \frac{2.25}{\lambda^2} = 4 \implies \lambda^2 = \frac{2.25}{4} = \frac{9}{16}$$ 46. So $\lambda = \pm \frac{3}{4}$. 47. For $\lambda = \frac{3}{4}$: $$x = \frac{1}{3/4} = \frac{4}{3}, y = \frac{1}{2 \cdot 3/4} = \frac{2}{3}, z = -\frac{4}{3}$$ 48. Evaluate $f$: $$f = 2 \cdot \frac{4}{3} + \frac{2}{3} - 2 \cdot \left(-\frac{4}{3} \right) = \frac{8}{3} + \frac{2}{3} + \frac{8}{3} = \frac{18}{3} = 6$$ 49. For $\lambda = -\frac{3}{4}$, signs reverse, $f = -6$. 50. Max 6 at $(\frac{4}{3}, \frac{2}{3}, -\frac{4}{3})$, min -6 at $(-\frac{4}{3}, -\frac{2}{3}, \frac{4}{3})$. --- 51. **Problem 6:** $f = xyz$, constraint $x^2 + y^2 + z^2 = 1$. 52. Gradients: $$\nabla f = (yz, xz, xy), \quad \nabla g = (2x, 2y, 2z)$$ 53. Set: $$yz = 2\lambda x, \quad xz = 2\lambda y, \quad xy = 2\lambda z$$ 54. If any of $x,y,z=0$, then $f=0$. 55. Assume none zero, divide equations pairwise: $$\frac{yz}{xz} = \frac{y}{x} = \frac{2\lambda x}{2\lambda y} = \frac{x}{y} \implies y^2 = x^2$$ Similarly, $x^2 = z^2$, $y^2 = z^2$. 56. So $|x|=|y|=|z|$. 57. Let $x = y = z = t$, then constraint: $$3t^2 = 1 \implies t = \pm \frac{1}{\sqrt{3}}$$ 58. Evaluate $f$: $$f = t \cdot t \cdot t = t^3 = \pm \frac{1}{3\sqrt{3}}$$ 59. Max $\frac{1}{3\sqrt{3}}$, min $-\frac{1}{3\sqrt{3}}$ at points with all coordinates equal in magnitude and sign. --- 60. **Problem 7:** $f = 3x + 6y + 2z$, constraint $2x^2 + 4y^2 + z^2 = 70$. 61. Gradients: $$\nabla f = (3,6,2), \quad \nabla g = (4x, 8y, 2z)$$ 62. Set: $$3 = \lambda 4x, \quad 6 = \lambda 8y, \quad 2 = \lambda 2z$$ 63. Solve: $$x = \frac{3}{4\lambda}, \quad y = \frac{6}{8\lambda} = \frac{3}{4\lambda}, \quad z = \frac{2}{2\lambda} = \frac{1}{\lambda}$$ 64. Substitute into constraint: $$2 \left( \frac{3}{4\lambda} \right)^2 + 4 \left( \frac{3}{4\lambda} \right)^2 + \left( \frac{1}{\lambda} \right)^2 = 70$$ 65. Calculate: $$2 \cdot \frac{9}{16\lambda^2} + 4 \cdot \frac{9}{16\lambda^2} + \frac{1}{\lambda^2} = 70$$ $$\frac{18}{16\lambda^2} + \frac{36}{16\lambda^2} + \frac{1}{\lambda^2} = 70$$ $$\frac{54}{16\lambda^2} + \frac{1}{\lambda^2} = 70$$ $$\frac{54 + 16}{16\lambda^2} = 70 \implies \frac{70}{16\lambda^2} = 70$$ 66. Multiply both sides by $16\lambda^2$: $$70 = 70 \cdot 16 \lambda^2 \implies 1 = 16 \lambda^2 \implies \lambda^2 = \frac{1}{16}$$ 67. So $\lambda = \pm \frac{1}{4}$. 68. For $\lambda = \frac{1}{4}$: $$x = \frac{3}{4 \cdot 1/4} = 3, y = 3, z = 4$$ 69. Evaluate $f$: $$f = 3 \cdot 3 + 6 \cdot 3 + 2 \cdot 4 = 9 + 18 + 8 = 35$$ 70. For $\lambda = -\frac{1}{4}$, $f = -35$. 71. Max 35 at $(3,3,4)$, min -35 at $(-3,-3,-4)$. --- 72. **Problem 8:** $f = x^4 + y^4 + z^4$, constraint $x^2 + y^2 + z^2 = 1$. 73. Gradients: $$\nabla f = (4x^3, 4y^3, 4z^3), \quad \nabla g = (2x, 2y, 2z)$$ 74. Set: $$4x^3 = \lambda 2x, \quad 4y^3 = \lambda 2y, \quad 4z^3 = \lambda 2z$$ 75. For $x \neq 0$: $$4x^3 = 2\lambda x \implies 2x^2 = \lambda$$ Similarly for $y,z$. 76. So $2x^2 = 2y^2 = 2z^2 = \lambda$, meaning $x^2 = y^2 = z^2$. 77. Let $x^2 = y^2 = z^2 = t^2$, then constraint: $$3t^2 = 1 \implies t^2 = \frac{1}{3}$$ 78. Evaluate $f$: $$f = 3t^4 = 3 \left( \frac{1}{3} \right)^2 = 3 \cdot \frac{1}{9} = \frac{1}{3}$$ 79. Check points where two variables zero and one is $\pm 1$: $$f = 1^4 = 1$$ 80. So max is 1 at $(\pm 1,0,0)$ and permutations, min is $\frac{1}{3}$ at points with equal magnitudes. --- **Summary:** Each problem solved using Lagrange multipliers with detailed steps.