A-LevelMathematicsDifferentiation9709/12 · Oct/Nov 2013 · Q67 Marks

In the diagram, S is the point (0, 12) and T is the point (16, 0). The point Q lies on ST, between S and T, and has coordinates (x, y). The points P and R lie on the x-axis and y-axis respectively and OPQR is a rectangle.

(i)Show that the area, A, of the rectangle OPQR is given by A = 12x – ¾x².[3]
(ii)Given that x can vary, find the stationary value of A and determine its nature.[4]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2013). The mark or mark type is shown in brackets.

[1] Sim triangles y/(16-x) = 12/16 (or trig) Trig, similarity or eqn of line (could also come from eqn of line)
[1] → y = 12 − ¾x
[1] A = xy = 12x – ¾x². ag - check working.
[1] dA/dx = 12 − 6x/4
[1] = 0 when x = 8. → A = 48. Sets to 0 + solution.
[1] This is a Maximum. Can be deduced without any working.
[1] From -ve quadratic or 2nd differential. Allow even if '48' incorrect.

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