Hint
The cut x must be less than half the shorter sheet side. Keep both derivative roots until you have checked this constraint.
Worked solution
(a)
The box dimensions are x, 24−2x and 18−2x. Thus \(\displaystyle V=x(24-2x)(18-2x)=432x-84x^{2}+4x^{3}\). Strict positivity gives x>0, x<12 and x<9, hence 0<x<9.
Answer: \(V=432x-84x^2+4x^3\) \(\displaystyle \mathrm{cm}^{3}\); \(0<x<9\).
(b)
\(\displaystyle V'=432-168x+12x^{2}=12(x^{2}-14x+36)\). Solving gives \(\displaystyle x= \frac{14\pm \sqrt{52}}{2}= 7\pm \sqrt{13}\). The plus value is about 10.606>9 and is infeasible; \(\displaystyle 7-\sqrt{13}\approx 3.39445\) lies in (0,9).
Answer: \(x=7-\sqrt{13}\) cm; reject \(7+\sqrt{13}>9\).
(c)
Let \(\displaystyle \alpha =7-\sqrt{13}\) and \(\displaystyle \beta =7+\sqrt{13}. V'=12(x-\alpha )(x-\beta )\). On the feasible interval β>9, so V′>0 for 0<x<α and V′<0 for α<x<9. Thus α gives the global maximum; the boundary volume limits are zero. Substitution gives \(V(\alpha)=(7-\sqrt{13})(10+2\sqrt{13})(4+2\sqrt{13})=280+104\sqrt{13}\) \(\displaystyle \mathrm{cm}^{3}\). This is approximately \(\displaystyle 654.977 \mathrm{cm}^{3}\), or \(\displaystyle 655 \mathrm{cm}^{3}\) to 3 significant figures.
Answer: \(280+104\sqrt{13}\) \(\displaystyle \mathrm{cm}^{3}\), approximately \(\displaystyle 655 \mathrm{cm}^{3}\).
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 3 marks
- 1 mark: Use all three correct dimensions in the volume model.
- 1 mark: Expand to \(\displaystyle 432x-84x^{2}+4x^{3}\).
- 1 mark: State the complete feasible domain 0<x<9.
Alternative methods: The factored volume is accepted alongside a correct polynomial expansion and domain.
Common error: Using x<12 alone and overlooking the shorter side.
(b) · 3 marks
- 1 mark: Differentiate and obtain the correct stationary quadratic.
- 1 mark: Find both exact roots \(\displaystyle 7\pm \sqrt{13}\).
- 1 mark: Retain only \(\displaystyle 7-\sqrt{13}\) using the feasible domain.
Alternative methods: Completing the square in the derivative equation is accepted.
Common error: Keeping an algebraic stationary root that produces a negative box width.
(c) · 3 marks
- 1 mark: Justify the global maximum over 0<x<9 by derivative signs or a complete candidate/boundary comparison.
- 1 mark: Evaluate the exact greatest volume \(\displaystyle 280+104\sqrt{13} \mathrm{cm}^{3}\).
- 1 mark: Give \(\displaystyle 655 \mathrm{cm}^{3}\) to 3 significant figures.
Alternative methods: A valid local classification plus complete comparison with boundary limits is accepted for global optimality; equivalent exact volume forms are accepted.
Common error: Using only V″<0 to claim a global result without domain reasoning, or rounding x too early.
Total: 9 marks.