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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
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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?
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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?
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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)
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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
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To simplify 400 – 160x + 12x2 = 0, divide by 4 100 – 40x + 3x2 = 0
Think aboutWhy not?
Think aboutWhy not?
Maximum volume of a box
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
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Maximum volume of a box
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?
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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
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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
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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 =
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Think aboutWhy is r = -4.30
not included here?
Think aboutWhy is r = -4.30
not included here?
Minimum material to make a can
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
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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?
Minimum material to make a can
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)
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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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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
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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?
Think aboutWhat do you do?
Think aboutWhat do you do?
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
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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