1. Problem A: Find intervals of increase/decrease and local extrema for $y = x^3 - 2x^2 + x - 3$. 2. Use the first derivative test: $y' = 3x^2 - 4x + 1$. 3. Solve $y' = 0$ for critical points: $3x^2 - 4x + 1 = 0$. 4. Using quadratic formula: $x = \frac{4 \pm \sqrt{16 - 12}}{6} = \frac{4 \pm 2}{6}$, so $x=1$ and $x=\frac{1}{3}$. 5. Test intervals around critical points: - For $x < \frac{1}{3}$, pick $x=0$: $y'(0)=1>0$ increasing. - Between $\frac{1}{3}$ and $1$, pick $x=0.5$: $y'(0.5) = 3(0.25) - 4(0.5) + 1 = 0.75 - 2 + 1 = -0.25 < 0$ decreasing. - For $x > 1$, pick $x=2$: $y'(2) = 12 - 8 + 1 = 5 > 0$ increasing. 6. Local max at $x=\frac{1}{3}$, local min at $x=1$. 7. Find $y$ values: - $y(\frac{1}{3}) = (\frac{1}{3})^3 - 2(\frac{1}{3})^2 + \frac{1}{3} - 3 = \frac{1}{27} - \frac{2}{9} + \frac{1}{3} - 3 = -\frac{71}{27}$. - $y(1) = 1 - 2 + 1 - 3 = -3$. --- 8. Problem B: Concavity and inflection for $y = x^3 - 12x^2 - 3x - 2$. 9. Second derivative: $y'' = 6x - 24$. 10. Set $y''=0$: $6x - 24 = 0 \Rightarrow x=4$ (inflection point). 11. Test concavity: - For $x<4$, $y''<0$ concave down. - For $x>4$, $y''>0$ concave up. 12. Find $y(4) = 64 - 192 - 12 - 2 = -142$. Inflection point at $(4, -142)$. --- 13. Problem C: Position $s(t) = t^3 - t^2 - 16t + 8$. 14. Velocity $v(t) = s'(t) = 3t^2 - 2t - 16$. 15. Set $v(t)=0$: $3t^2 - 2t - 16=0$. 16. Solve quadratic: $t = \frac{2 \pm \sqrt{4 + 192}}{6} = \frac{2 \pm 14}{6}$. 17. Solutions: $t=\frac{16}{6} = \frac{8}{3}$ and $t=\frac{-12}{6} = -2$ (discard negative time if context). 18. Acceleration $a(t) = v'(t) = 6t - 2$. 19. At $t=\frac{8}{3}$, $a = 6(\frac{8}{3}) - 2 = 16 - 2 = 14$ m/s². --- 20. Problem D: Maximize $(x+1)(y+2)$ given $x + 2y = 50$. 21. Express $y = \frac{50 - x}{2}$. 22. Substitute: $f(x) = (x+1)(\frac{50 - x}{2} + 2) = (x+1)(\frac{50 - x + 4}{2}) = \frac{(x+1)(54 - x)}{2}$. 23. Expand: $f(x) = \frac{-x^2 + 53x + 54}{2}$. 24. Derivative: $f'(x) = \frac{-2x + 53}{2} = -x + \frac{53}{2}$. 25. Set $f'(x)=0$: $x = \frac{53}{2} = 26.5$. 26. Second derivative $f''(x) = -1 < 0$ confirms maximum. 27. Find $y$: $y = \frac{50 - 26.5}{2} = \frac{23.5}{2} = 11.75$. --- 28. Problem F: Open box from 30 in by 14 in cardboard by cutting squares of side $x$. 29. Volume $V = x(30 - 2x)(14 - 2x) = x(420 - 60x - 28x + 4x^2) = x(420 - 88x + 4x^2)$. 30. Expand: $V = 420x - 88x^2 + 4x^3$. 31. Derivative: $V' = 420 - 176x + 12x^2$. 32. Set $V' = 0$: $12x^2 - 176x + 420 = 0$. 33. Divide by 4: $3x^2 - 44x + 105 = 0$. 34. Solve quadratic: $x = \frac{44 \pm \sqrt{44^2 - 4 \cdot 3 \cdot 105}}{6} = \frac{44 \pm \sqrt{1936}}{6} = \frac{44 \pm 44}{6}$. 35. Solutions: $x=0$ or $x=\frac{88}{6} = 14.67$ (too large, discard). 36. Check critical points and endpoints $x=0$, $x=7$ (max possible cut). 37. Evaluate $V$ at $x=0$: $0$; at $x=7$: $7(30-14)(14-14)=0$. 38. Since $V'$ only zero at $x=0$ in domain, check $V$ at $x=3$: $V(3) = 3(30-6)(14-6) = 3(24)(8) = 576$. 39. By testing values, max volume at $x=3$ inches. 40. Justification: $V'' = -176 + 24x$, at $x=3$, $V'' = -176 + 72 = -104 < 0$ confirms max. 41. Maximum volume is $576$ cubic inches.