Nuffield Free-Standing Mathematics Activity Maxima and minima

1. Nuffield Free-Standing Mathematics Activity Maxima and minima

2. Maxima and minima What is the maximum area that can be enclosed with 200 metres of fence? What is the minimum area of aluminium sheet that can be used to make a can to hold 330 ml? What are the dimensions of the box with the greatest volume that can be made from a 150 cm by 120 cm sheet of cardboard? This activity is about using calculus to solve problems like these.

3. = nxn – 1 = naxn – 1 = m = 0 Rules of differentiation Function Derivative y = xn y = axn y = mx y = c

4. = 0 = 0 is negative At a minimum point At a point of inflexion = 0 is positive Maxima and minima At a maximum point

5. 10 –x(cm) For maximum area = 0 x (cm) Example A piece of wire 20 cm long is bent into the shape of a rectangle. What is the maximum area it can enclose? A = x(10 – x ) =10x – x2 – 2x = 10 2x= 10 x = 5 =– 2 implies a maximum The area is maximum when the shape is a square. Maximum area A =25 cm2

6. v v= 3t – 0.2t2 11.25 = 0 t= 0 t 7.5 15 Example The velocity of a car, v m s–1, between 2 sets of traffic lightsis modelled by v = 3t – 0.2t2 where t is the time in seconds. =3 – 0.4t This gives acceleration. The maximum speed occurs when: 0.4t= 3 =7.5 (s) =– 0.4 implies a maximum Maximum speed v = 3  7.5 – 0.2  7.52 =11.25 (m s–1)

7. p = x3 –18x2 + 105x– 88 maximum (5, 112) p 0 x – 88 minimum (7, 108) Example The functionp= x3 – 18x2 + 105x– 88gives the profit p pence per item when x thousand are produced. = 3x2 –36x = 0 for maximum/minimum + 105 x2 – 12x+ 35 = 0 x= (x )(x ) = 0 –5–7 5 or 7 =6x – 36 When x = 5, this is negative When x = 7, it is positive Maximump = 53 – 18  52 + 105 5– 88 =112 (pence) Minimump = 73 – 18  72 + 105 7 – 88 =108 (pence)

8. radiusr metres heighth metres Capacity4 m3 h = 2r2+ 2rh S = Example Find the radius and height thatgive the minimum surface area. S = 2r2+ 8r–1 = 4r – 8r–2 = 0 for max/min 4 r V = r2h= 4 S = 2r2+ 2rh r3 =0.6366... S = 2r2+ 2r 0.860... (m) r = + 16r–3 positive-minimum =4 Radius 0.86m and height 1.72m (2dp) gives the minimum surface area. =1.72... (m) =13.9 (m2)

9. Maxima and minima • Reflect on your work Explain how you set about constructing a formula from a situation. What are the steps in the method for finding the maximum or minimum value of the function?