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 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.

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

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

5. 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

6. 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
= (m s–1)

7. Example
The function p = x3 – 18x x – 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 –18x x – 88
– 88
maximum
(5, 112)
(x )(x ) = 0
– – 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   5 – 88
= 112 (pence)
Minimum p = 73 – 18   7 – 88
= 108 (pence)

8. 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 =
S = 2r2 + 2rh
r =
(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 =
= (m)
S =
2r2 + 2rh
= 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?