By Stan Goldberg and Bob Smith

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**Example text**

Is a maximum point. 1 2 3 4 (ii) y = 2 x + x 2 − 4 x 3 = x(2 + x − 4 x 2 ) = − x(4 x 2 − x − 2) The curve cuts the x-axis at x = 0 and at the points satisfying 4x 2 − x − 2 = 0 . For this quadratic equation, a = 4,b = −1, c = −2 1 ± 1 − 4 × 4 × −2 1 ± 33 Using the quadratic formula, x = = 8 8 ( 21 , 43 ) 1 8 (1 − 33 ) (− 1 3 ,− 11 27 ) 1 8 (1 + 33 ) © MEI, 11/08/08 2/7 MEI C2 Differentiation Section 2 Exercise solutions 3. y = x 3 − 3 x 2 + 6 dy dx = 3x 2 − 6x At turning points, dy =0 dx 3x 2 − 6x = 0 3 x( x − 2) = 0 x = 0 or x = 2 When x = 0, y = 6 When x = 2, y = 2 3 − 3 × 2 2 + 6 = 8 − 12 + 6 = 2 The turning points are (0, 6) and (2, 2).

I) 24 x x 24 - 2x x 15 - 2x 15 x Height of box is x cm Length of box is (24 – 2x) cm Width of box is (15 – 2x) cm Volume V = x(15 − 2 x )(24 − 2 x ) = x(360 − 78 x + 4 x 2 ) = 4 x 3 − 78 x 2 + 360 x dV = 12 x 2 − 156 x + 360 dx At turning points, 12 x 2 − 156 x + 360 = 0 (ii) x 2 − 13 x + 30 = 0 ( x − 3)( x − 10) = 0 x = 3 or x = 10 © MEI, 11/08/08 6/7 MEI C2 Differentiation Section 2 Exercise solutions x = 10 is not possible since this would mean that the width would be negative. d2V = 24 x − 156 dx 2 d2V When x = 3, = 72 − 156 < 0 , so x = 1 is a maximum point.

Dt dt 1 2 After 2 seconds, the velocity v m s and acceleration a m s are given by: (a) v = 30 and a = 12 (c) v = 12 and a = 6 (e) I don’t know (b) v = 6 and a = 0 (d) v = 6 and a = 12 6. Given that x + y = 60, the maximum value of x2y is (a) 40 (c) 16000 (e) I don’t know (b) 32000 (d) 20 Questions 7 and 8 are about a straight wall AB and a fence of length 5 m which form a rectangular enclosure. The width of the enclosure is x m. A B xm 7. The area of the enclosure, in m², is given by (a) x(5 x) (c) x(5 2 x) (e) I don’t know (b) x(2 x 5) (d) x( x 5) 8.

### Archie Vol 1 No 555 May 2005. by Stan Goldberg and Bob Smith

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