Free-Standing Mathematics Activity
Maximising
and
minimising
© Nuffield Foundation 2011
Manufacturers use containers of different shapes and sizes
How can they design containers to:
• hold as much as possible
• use as little material as possible?
In this activity you will use calculus to solve such problems
© Nuffield Foundation 2011
Maximum volume of a box
Step 1 – find a formula
A piece of card measures
20 cm by 20 cm
A square with sides of length
x cm is removed from each
corner
An open-topped box is made
by folding the card
© Nuffield Foundation 2011
Think about
What are the dimensions
of the box?
Can you find an expression
for the volume of the box?
Maximum volume of a box
Volume of the box
V = x(20 – 2x)(20 – 2x)
V = x(400 – 80x + 4x2)
V = 400x – 80x2 + 4x3
Think about
What must be done next to find the
maximum value of V?
How can you be sure this gives the
maximum?
What else could you do?
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20 – 2x
20
Maximum volume of a box
Step 2 – use calculus
Find the first and second derivatives
Solve the equation
dV
dx
=0
This gives the values of x for which the function V
has turning points
Then work out the maximum value of V
Use the second derivative to check that this is a
maximum (and not a minimum)

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Maximum volume of a box – step 2
V = 400x – 80x2 + 4x3
Let V = V(x), then the first derivative = V ′(x)
and the second derivative = V ″(x)
V ′(x) = 400 – 160x + 12x2
V ″(x) = -160 + 24x
V has turning points where V ′(x) = 0
So 400 – 160x + 12x2 = 0
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Maximum volume of a box
To simplify 400 – 160x + 12x2 = 0, divide by 4
100 – 40x + 3x2 = 0
Think about
Would you solve this using
factors or the formula?
The determinant, b2 – 4ac = (-40)2 – 4 × 3 × 100
= 1600 – 1200 = 400 = 202
A square number means there are factors:
100 – 40x + 3x2 = (10 – x)(10 – 3x) = 0
So x = 10 or
But x = 10 is not a reasonable solution
So x =
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is the only one that can apply
Think about
Why not?
Maximum volume of a box
Does x =
give a maximum?
V″(x) = -160 + 24x
When x =
V″ (
) = -160 + 24 ×
= -80 < 0
So this does give a maximum
V max = 400( ) – 80( )2 + 4(
V max = 593 cm3 (3 sf)
© Nuffield Foundation 2011
)3
What if you wanted the minimum material to
make a cylinder with a required volume?
In this case you would have two
variables (radius and height) and
one fixed quantity (volume)
Think about
Why is having two
variables a problem?
In order to differentiate, you need an expression for the
quantity you want to minimise (or maximise) in terms of
just one variable
© Nuffield Foundation 2011
Working with a cylinder
First, use the fixed volume to eliminate one of the
variables (either the height or radius)
When you have an expression for the quantity of
material needed to make the cylinder in terms of just
one variable, differentiate it and put the derivative = 0
Solve this equation to find the value of the variable that
gives a minimum (or maximum)
Then find the value of the other variable and the
minimum (or maximum) that you require
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Minimum material to make a can
Say you want to find the minimum metal needed to
make a can to hold 500 ml (the same as 500 cm3)
If r cm is the radius and h cm is the height,
the volume V = πr2h
and the metal used M = 2πr2 + 2πrh
So if V = 500 then πr2h = 500
It is easier to eliminate h
(because r is a squared term)
h=
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Think about
Why is this the
area of metal
needed?
Think about
Which variable is it
easier to get rid of?
Minimum material to make a can
Using h =
M = 2πr2 + 2πrh = 2πr2 + 2πr ×
M = 2πr2 + 2π × 500 = 2πr2 + 1000r-1
πr
dM
So = dr
= 4πr − 1000r-2
When
dM
=
dr
0 there will be turning point
So 4πr − 1000r-2 = 0 giving 4πr = 1000r-2
4πr3 = 1000
r3 = 250
π
r = 4.30 (3 sf)
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Think about
Why is r = -4.30
not included here?
Minimum material to make a can
r = 4.30 (3 sf)
Does this give a minimum?
M ′(r) = 4πr − 1000r-2 so M ″(r) = 4π + 2000r-3
When r = 4.30, M ″(r) is positive
So this will give a minimum value for M
Now h =
= 500 π (π × 4.302) = 8.60 (3 sf)
So M = 2πr2 + 2πrh = 2π × 4.302 + 2π × 4.30 × 8.60
= 349 cm2 (3 sf)
Think about
If r were eliminated instead of h,
would this answer be the same?
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Now set your own problem
Solve a packaging problem that needs:
• either to use the least materials
• or to hold the most volume
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You could adapt one of the examples
With a can, you might decide that the material used
in the base and top needs to be double thickness,
so you would end up with a different answer
or you might decide that two different metals
should be used with a different unit cost for each
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There are many other possibilities
Boxes come in all sorts of shapes, with and without lids −
what about a Toblerone box?
Or how about swimming pools, with a shallow end and a
deep end? Would the cement be equally thick all over?
What about ice cream cone packaging?
What about a wooden play house? (Remember the door!)
http://mathforum.org/dr.math/faq/formulas/
gives lots of other formulae
© Nuffield Foundation 2011
Summary of method
Think about
What do you do?
If you have only one variable and one given value,
use these to create a formula for the term that is to be
minimised/maximised
Think about
What do you do?
If you have two variables and one given value,
first decide how to write one of the variables in terms of
the other variable and the given value
Only then create the formula for
minimisation/maximisation
© Nuffield Foundation 2011
Summary of method
Think about
What do you
do next?
Now you have an expression for the quantity
you want to maximise/minimise
Differentiate it with respect to the variable, put the result
equal to 0 and solve the equation you get
To ensure you have a valid answer to the problem, find
the second derivative of the expression
Substitute each value of the variable to see which, if any,
gives the required maximum/minimum
Finally work out the dimensions you need
and then the quantity you wanted to
maximise/minimise
© Nuffield Foundation 2011
Think about
What is the rule
that tells you
which you have?