Transcript Climate Change - Nuffield Foundation

Nuffield Free-Standing Mathematics Activity
Maxima and minima
© Nuffield Foundation 2011
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 = xn
dy
dx
= nx n – 1
y = ax n
dy
dx
= nax n – 1
y = mx
dy
dx
=m
y=c
dy
dx
=0
Maxima and minima
At a maximum point
dy
dx
=0
d2 y
is negative
dx 2
At a minimum point
dy
dx
=0
d2 y is positive
dx 2
At a point of inflexion
d2 y
dx 2
=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
dA
dx
10 – x
(cm)
= 10 – 2x
For maximum area
2x = 10
dA
dx
=0
x=5
x (cm)
d2 A
dx2 = – 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.
dv
dt
= 3 – 0.4t This gives acceleration.
The maximum speed occurs when:
dv
dt
=0
v
11.25
0.4t = 3
v = 3t – 0.2t2
3
t = 0.4 = 7.5 (s)
d2v
dt 2 = – 0.4 implies a maximum
Maximum speed v = 3  7.5 – 0.2  7.52
= 11.25 (m s–1)
0
7.5
15 t
Example
The function p = x3 – 18x2 + 105x – 88 gives the profit
p pence per item when x thousand are produced.
dp
dx
= 3x2 – 36x + 105 = 0 for maximum/minimum
p = x3 –18x2 + 105x – 88
x2 – 12x + 35 = 0
(x – 5 )(x – 7) = 0
x = 5 or 7
d2 p
dx 2 = 6x – 36
p maximum
(5, 112)
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)
0
– 88
x
minimum
(7, 108)
Example
Find the radius and height that
give the minimum surface area.
S = 2r2 + 8r–1
dS
dr
= 4r – 8r–2 = 0 for max/min
4 r  8
r2
r 3  8 = 0.6366...
4
r = 0.860... (m)
d2 S
–3 positive-minimum
+
16r
=
4
2
dr
h=
4
 r 2
= 1.72... (m)
S = 2r2 + 2rh = 13.9 (m2)
radius
r metres
Capacity height
4 m3
h metres
V = r2h = 4
S = 2r2 + 2rh
S = 2r2 + 2r  πr4 2
Radius 0.86m and
height 1.72m (2dp)
gives the minimum
surface area.
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?