O-LevelAdditional MathematicsDifferentiation and integration4037/21 · May/June 2013 · Q56 Marks
A piece of wire of length 96cm is formed into the rectangular shape PQRSTU shown in the diagram. It is given that PQ = TU = SR = xcm. It may be assumed that PQ and TU coincide and that TS and QR have the same length. [Figure 5.1]
- (i)Show that the area, A cm², enclosed by the wire is given by A = (96x - 3x²) / 2.[2]
- (ii)Given that x can vary, find the stationary value of A and determine the nature of this stationary value.[4]
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Mark scheme
Marking points from the official mark scheme (May/June 2013). The mark or mark type is shown in brackets.
- [1] (i) (QR = PS =) (96-3x)/2
- [1] (i) Area = x(96-3x)/2 (AG)
- [1] (ii) dA/dx = (96-6x)/2 or 48-3x o.e.
- [1] (ii) Solving dA/dx = 0
- [1] (ii) x = 16
- [1] (ii) A = 384 and state maximum
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