2020 RI P1 Q11

2020 RI P1 Q11

12 marks
Prelims

Figure 1 shows an open container in the form of a trapezoidal prism \(ABCDEFGH\) with square base \(ABCD\) and \(AB=AE=BF=EH=a\)cm, where \(a\) is a constant. The container is made of plastic of negligible thickness and is placed on a horizontal surface. The faces \(BCGF\) and \(ADHE\) are inclined at an angle \(\theta \) radians, \(0<\theta <\frac{\pi }{2}\), to the horizontal, and faces \(ABFE\) and \(DCGH\) are perpendicular to base \(ABCD\) . Figure 2 shows its cross-sectional view.

  1. Show that the volume \(V\)cm\(^{3}\) of the container is given by \(V={{a}^{3}}\sin \theta \left( 1+\cos \theta \right)\).[2]
  2. Use differentiation to find, in terms of \(a\), the maximum value of \(V\) in exact form, proving that it is a maximum.[5]
  3. A container is constructed with \(\theta =\frac{\pi }{3}\) and it is filled with water to half its height. Find, in terms of \(a\), the exact volume of water in the container.[3]
    The container is then tilted in the direction of the face \(BCGF\) until face \(BCGF\) and base \(ABCD\) makes the same angle \(\beta \) radians, where \(\beta \) is a constant, with the horizontal. Figure 3 shows its cross-sectional view.
    List item image
    Explain if it is possible to tilt the container to this position without any water flowing out from the container.[2]

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Answer:(i) V = a³sinθ(1+cosθ) (ii) θ = π/3, max V = (3√3/4)a³ (iii) Not possible, half-fill volume exceeds inverted capacity

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