Maximum volume/space or
minimum materials
Design your own assignment
Let’s look at a couple of ideas
Boxes 1
• A piece of card is 20cm
by 20cm
• The side of the squares
removed from each
corner is x cm
• A tray is made by
folding the card
• Find the maximum
volume, V cm3, that can
be held inside the tray
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Boxes 1 (solution)
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Volume of the box
V = x(20 – 2x)(20 – 2x)
V = x(400 – 80x + 4x2)
V = 400x – 80x2 + 4x3
20 – 2x
20
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Boxes 2
• Find dV/dx and the 2nd derivative
• Solve the equation when dV/dx = 0
• Hence find the values of x for which
the function V has turning points
• and establish the maximum volume for
the box that can be made from the
sheet of card
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Boxes 2 (solution 1)
V = 400x – 80x2 + 4x3
Let V = V(x),
So dV/dx = V ’(x)
and the 2nd derivative = V ’’(x)
V ’(x) = 400 – 160x + 12x2
V ’’(x) = -160 + 24x
V will have turning points where V ’(x) = 0
So 400 – 160x + 12x2 = 0
100 – 40x + 3x2 = 0
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Boxes 2 (solution 2)
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100 – 40x + 3x2 = 0
Is this solved by formula or factorisation?
The determinant, b2 – 4ac = (-40)2 – 4.3.(100)
= 1600 – 1200 = 400 = 202, a square number
So factorise:
100 – 40x + 3x2 = (10 – x)(10 – 3x) = 0
x = 10 or 10/3
But given x < 10 (half the width) x = 10 is not a
reasonable solution for this situation
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Boxes 2 (solution 3)
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x = 10/3 – is it a maximum?
V’’(x) = -160 + 24x
When x = 10/3
V’’(10/3) = - 160 + 24 x 10/3 < 0
 maximum
Vmax = 400(10/3) – 80(10/3)2 + 4(10/3)3
Vmax = 593 cm3 (3 sf)
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Start to think about assignments
• Devise a packaging problem that will
either use the least materials or hold the
most volume
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Start to think about assignments
• It could be a tin can, in which case you
would have 2 variables (radius and height)
+ one fixed quantity (volume)
• So you would find an expression for
calculating the fixed quantity and then
use this expression to get rid of either
the height or the radius
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Start to think about assignments
• Then form an expression for the quantity
you want to make as small (or as large) as
possible
• Differentiate it, put = 0 and find the
value of x that will give the minimum (or
maximum)
• Hence find the other dimension and the
value of the material or volume that you
require
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Let me demonstrate
• Say I want to use the minimum metal to
make a can to hold 500ml ( which is the
same as 500 cm3)
• If r cm is the radius and h cm is the
height, the volume V = r2h
• and the metal used M = 2r2 + 2rh
• So if V = 500 then r2h = 500
• I will eliminate h (its easier than a
squared term) so h = 500/ r2
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Let me demonstrate
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Using h = 500/ r2
M = 2r2 + 2rh = 2r2 + 2r x 500/ r2
M = 2r2 + 2 x 500/ r = 2r2 + 1000r-1
So dM/dr = 4r - 1000r-2
When dM/dr = 0 we will have turning point
So 4r - 1000r-2 = 0 4r = 1000r-2
4r3 = 1000  r3 = 250/   r = 4.30(3 sf)
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Let me demonstrate
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r = 4.30(3 sf)
Is this a minimum?
M’’(x) = 4 + 2000r-3
At r = 4.30
M’’(4.30) = 4 + 2000/(4.30)3 > 0  a minimum
So h = 500  (18.5) =8.60 (3sf)
M= 2  x 18.5 + 2 x 4.3 x 8.6 = 349 cm2 (3 sf)
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Don’t think you can do that one!
• If you did a can, I would expect for
instance that you would decide the base
and the top would be double thickness, so
you would end up with different answers
• or that 2 different metals would be used
with a different unit cost for each
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But there are dozens of others
• Boxes of all 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!)
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Some useful formulae
A cone
• Volume: 1/3 r2h
• Surface area :
• 2r is the length of the circle
at the top of the cone
• The curved surface of the cone
comes from a circle radius 2s
• So area of circle radius s is s2
and the cone has r/s of it
• So area = s2 x r/s = sr
• where s = (h2 + r2)
r
h
s
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Some useful formulae
Part of a sphere (upside down!)
• Radius of sphere: r
Radius of base: r1
Height: h
Surface area: S
Volume: V
•
r = (h2+r12)/(2h)
S = 2 rh
V = (/6)(3r12+h2)h
• http://mathforum.org/dr.math/faq/for
mulas/ for lots of other formulae
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