The town council in a HDB housing estate is planning to construct a shed to store gardening equipment for a community garden.
The shed is made up of three parts.
The roof is modelled by the four identical triangular surfaces of a right pyramid. This right pyramid has a square base with sides \(2x\) metres and a height of \(x\) metres.
The four vertical walls are modelled by rectangles, each of sides \(2x\) metres by \(h\) metres.
The floor is modelled by a square of sides \(2x\) metres.
The three parts are joined together as shown in the diagram below. The shed is made of material of negligible thickness.
Due to budgeting, the material used to build this shed must be exactly \(72\ \mathrm{m}^2\) of treated plywood (including the flooring). To support the needs of the garden, the total volume of the shed must be at least \(36\ \mathrm{m}^3\).
By finding the height \(h\) in terms of \(x\), show that the volume \(V\) of the shed is given by \(V=36x-\left(\frac{2}{3}+2\sqrt{2}\right)x^3.\)[5]
Find the exact value of \(x\) that maximises the volume of the shed and determine whether the maximum volume meets the needs of the garden.[5]
It was decided that \(x=1.8\ \mathrm{m}\) for the actual construct of the shed. Due to wear and tear, a hole was formed at the top of the roof. During a heavy downpour, rainwater flows into the shed at a rate of \(1\ \mathrm{m}^3\mathrm{min}^{-1}\). Assuming that there is no leakage, determine the time taken to flood the shed up to the base of the pyramid-shaped roof.[2]
[Volume of right pyramid \(=\frac{1}{3}\times\)(area of base)\(\times\)(height of pyramid)]