Some positive numbers add up to 19. What is the maximum product?
With integers: 3  3  3  3  3  2  2  972 .
6
19
 19 
Not restricted to integers: Since 6   19 then    1008 is a possible
6
6
7
8
 19 
 19 
product as are    1085.398 and    1012 .
7
8
Assume all the numbers are the same. Let this number be
19
and so we have
x
x of them.
x
 19 
The product, P    . Taking natural logarithms of both sides we have
 x
x
 19 
 19 
ln P  ln    x ln    x  ln19  ln x 
 x
 x
If we want to find the maximum value of P as x changes we need to solve
dP
 0 . The problem is that we can differentiate the right hand side of
dx
ln P  x  ln19  ln x  but the left hand side looks awkward.
To see how to do it, let z  ln P .
Now
dz 1
dz dz dP 1 dP
.
 and by the chain rule



dP P
dx dP dx P dx
So differentiating z  ln P  x  ln19  ln x  with respect to x we get:
dz d
  x  ln19  ln x  
dx dx
1 dP
 1
  ln19  ln x   x   
P dx
 x
Making
dP
the subject and simplifying:
dx
dP
 P  ln19  ln x  1
dx
© MEI 2006
dP
 0 . Since we know the product P  0
dx
19
then we need to solve ln19  ln x  1  0 . This is equivalent to ln  1 and so
x
19
e.
we require
x
The maximum value occurs when
x
 19 
Therefore the maximum product is P     e e  1085.406 , slightly bigger
 x
19
7
 19 
than    1085.398 .
7
But there’s a problem. We were asked to split 19 up into ‘some numbers’.
Doesn’t this mean that we have to have, for example, 5 or 6 or 7 numbers
19
 6.9897 lots of e ? Can you really split 19 up into 6.9897 parts,
rather than
e
or even into 6 12 parts for that matter?
The sensible way to interpret the problem is that we do need a whole number
of numbers so the best way forward seems to be to use 6 lots of e and the
remaining number will then be 19  6e  2.69031 . The product of these seven
7
 19 
numbers is then e  19  6e   1085.348 but this is less than    1085.398 .
7
19
So does seven lots of
generate the maximum product?
7
6
© MEI 2006