Nuffield Free-Standing Mathematics Activity Maxima and minima

Maxima and minima What is the maximum area that can be enclosed with 200 metres of fence? 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? What is the minimum area of aluminium sheet that can be used to make a can to hold 330 ml? This activity is about using calculus to solve problems like these.

Rules of differentiation Function Derivative y = x n = nx n – 1 y = ax n = nax n – 1 y = mx = m = 0 y = c

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

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 x (cm) 10 – x (cm) = 10 – 2x For maximum area = 0 2x = 10 x = 5 = – 2 implies a maximum The area is maximum when the shape is a square. Maximum area A = 25 cm2

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

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

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

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?