Maximising and minimising

© Nuffield Foundation 2011

Free-Standing Mathematics Activity

Maximising and minimising


How can they design containers to:• hold as much as possible• use as little material as possible?

Manufacturers use containers of different shapes and sizes

In this activity you will use calculus to solve such problems


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

Think about What are the dimensions of the box?Can you find an expression for the volume of the box?

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 boxV = x(20 – 2x)(20 – 2x)V = x(400 – 80x + 4x2)V = 400x – 80x2 + 4x3

20

20 – 2x

Think aboutWhat 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?

Think aboutWhat 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?


Maximum volume of a box Step 2 – use calculus

Find the first and second derivatives

Solve the equation = 0

dVdx

This gives the values of x for which the function V has turning points Then work out the maximum value of VUse the second derivative to check that this is a maximum (and not a minimum)


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) = 0So 400 – 160x + 12x2 = 0

Maximum volume of a box – step 2


To simplify 400 – 160x + 12x2 = 0, divide by 4 100 – 40x + 3x2 = 0

Think aboutWhy not?

Think aboutWhy not?


Think aboutWould you solve this using factors or the formula?

Think aboutWould 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) = 0So x = 10 or But x = 10 is not a reasonable solution

So x = is the only one that can apply



When x =

V″ ( ) = -160 + 24 × = -80 < 0

V max = 400( ) – 80( )2 + 4( )3

V″(x) = -160 + 24x

So this does give a maximum

V max = 593 cm3 (3 sf)

Does x = give a maximum?


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?

Think aboutWhy 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


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

Working with a cylinder


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 + 2πrhThink about

Why is this the area of metal

needed?

Think aboutWhy is this the area of metal

needed?

Minimum material to make a can

Think aboutWhich variable is it easier to get rid of?

Think aboutWhich variable is it easier to get rid of?

So if V = 500 then πr2h = 500It is easier to eliminate h (because r is a squared term)

h =


Think aboutWhy is r = -4.30

not included here?

Think aboutWhy is r = -4.30

not included here?


Using h = M = 2πr2 + 2πrh = 2πr2 + 2πr ×

So 4πr − 1000r-2 = 0 giving 4πr = 1000r-2

4πr3 = 1000

r = 4.30 (3 sf)

M = 2πr2 + 2π × = 2πr2 + 1000r-1500πr

So = = 4πr − 1000r-2dM dr When = 0 there will be turning pointdM

dr

r3 = π250


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

Think aboutIf r were eliminated instead of h, would this answer be the same?

Think aboutIf r were eliminated instead of h, would this answer be the same?


Now h = = 500 π (π × 4.302) = 8.60 (3 sf)

So this will give a minimum value for M

So M = 2πr2 + 2πrh = 2π × 4.302 + 2π × 4.30 × 8.60 = 349 cm2 (3 sf)


Now set your own problem

Solve a packaging problem that needs:• either to use the least materials• or to hold the most volume


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


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!)

There are many other possibilities

http://mathforum.org/dr.math/faq/formulas/ gives lots of other formulae


If you have two variables and one given value,

Summary of method

If you have only one variable and one given value,

Think aboutWhat do you do?




use these to create a formula for the term that is to be minimised/maximised

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


Now you have an expression for the quantity you want to maximise/minimise

Think aboutWhat is the rule

that tells you which you have?

Think aboutWhat is the rule

that tells you which you have?

Think aboutWhat do you

do next?

Think aboutWhat do you

do next?

Summary of method

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/minimumFinally work out the dimensions you need and then the quantity you wanted to maximise/minimise

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Maximising and minimising