1785
/www/nrich/html/content/96/11/six6/
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2011-02-01T00:00:01
<mdoxml version="1.0"><p>Various sets of numbers add together to give a sum of $10$:</p>
<ul>
<li>$3 + 7 = 10$</li>
<li>$2 + 8 = 10$</li>
<li>$2.1 + 5.6 + 2.3 = 10$</li>
</ul>
<p>The products of these sets are all different:</p>
<ul>
<li>$3 \times7 = 21$</li>
<li>$2 \times 8 = 16$</li>
<li>$2.1 \times 5.6 \times2.3 = 27.048$</li>
</ul>
<p>What is the greatest product that can be made from numbers that add up to $10$?</p>
<p>Try using different starting numbers.</p>
<p>Can you find a strategy for splitting numbers so that you always get the largest product?</p>
<br></br>
<p><a href="http://nrich.maths.org/8010">Click here for a poster of this problem</a>.</p></mdoxml>
<mdoxml version="1.0"><br></br>
<p class="editorial">Thank you to all those who tried this problem,
there were a large number of solutions received. Luke and Alex from
Aqueduct Primary sent in their solution:</p>
<div>The largest product that we could find was $2.5 \times 2.5
\times 2.5 \times2.5 = 39.06$.</div>
<div>We spent hours looking at other numbers but realised that
larger numbers often made smaller products.</div>
<div>This was a really fun challenge!</div>
<br></br>
<div class="editorial"><br></br>
Matthew from Bydales realised that we have to exclude negatives,
otherwise solutions like $99\times - 44\times - 45 = 196 020 $ are
possible. He wrote:</div>
<div>Using negative numbers you can make the number as big as you
like as long as the other two numbers are negative and will leave
you with 10 when they are added. You can therefore get close to
inifinity if negatives are allowed!</div>
<p class="editorial"></p>
<p class="editorial">Thank you to all of the children at St
George's CE Primary School who had a go at this problem. Nathan,
Otis, Hannah and Leon all sent in the correct answer. Saif fom
Durston School, Sion and Daniel from TES and Cameron from Tokoroa
Intermediate in New Zealand all sent in correct solutions as
well.</p>
<p class="editorial"></p>
<p class="editorial">Andy from Garden International School sent in
a very clearly explained solution showing his working:</p>
<div>First we divided ten by $2$, then ten by $3$, then ten by $4$
and so on until we reach $9$.</div>
<div>Then we multiplied the divided number. In the case of $2$
is:</div>
<div>$5\times 5$</div>
<div>In the case of $3$ is: $3\frac{ 1}{3}\times3\frac{
1}{3}\times3\frac{ 1}{3}$.</div>
<div>We divided ten because the same number multiplied together
would give the largest product.</div>
<div>We started with two because there's no point using one and we
end with $9$ because there's no point multiplying $1$ by $1$ nine
times. In the end $2 \frac{1}{2} \times 2 \frac{1}{2} \times 2
\frac{1}{2} \times 2 \frac{1}{2}$ gives the largest product and the
product is $39.0625$.</div>
<p class="editorial"></p>
<p class="editorial">Kang Yun Seok, also from Garden International
School sent in a complete solution and discussed how similar
numbers give a larger product due to the properties of shapes like
squares and circles. This was also discussed in other solutions
including the one sent in by Mikey of Tadcaster School.</p>
<div>The largest product of any two numbers is from numbers that
are as similar as possible. That is why a $20 \times20$ square has
bigger area than a rectangle of the same perimeter - say, $39
\times1$ ...(that is because the biggest shape you can make with a
length of string is a circle or square instead of a long thin
rectangle).</div>
<div>So the largest product from ten is... with</div>
<div>$2$ numbers that add up to ten: $5 \times5$</div>
<div>$3$ numbers that add up to ten: $3.333 \times3.333
\times3.33$</div>
<div>4...</div>
<div>5...</div>
<div>and the largest of these products is: $2 \frac{1}{2} \times 2
\frac{1}{2} \times 2 \frac{1}{2} \times 2 \frac{1}{2}=
39.0625$.</div>
<br></br>
<div>It soon becomes apparent that when $10$ is split into $n$
parts (where $n$ is a whole number), the largest product is given
by $(\frac{10}{n})^n$ . So the problem is in fact to find the
largest value of $(\frac{10}{n})^ n$ . The largest value of
$(\frac{10}{n})^ n$ is when $n=4$ that is $39.0625$. This is the
largest product.</div>
<br></br>
<br></br>
<div><span class="editorial">Thomas Hu from A Y Jackson school used
his knowledge of the euler number ($e=2.71828...$) to find the
maximal solution for all values of $x$</span></div>
<p>Given the product $(\frac{x}{n})^n$ (where $x$ is the number in
question and $n$ in the number of parts it is being divided into).
It is already clear that repeated multiplication of the same number
$(2.5^4)$ is greater than that of two different numbers $(4*6)$ due
to maximization and difference of squares $(x-a)(x+a)=
x^2-a^2$.</p>
<p>Although the optimal $n$ for $10$ was stated to be $4$, that is
only true if one assumes that $n$ must be an integer. Otherwise, $n
= 3.7, 3.68, 3.679, 3.6788$ each provides increasingly larger
products. But these cannot work as the number must still add to the
value $x$.</p>
<p>But to maximise $(\frac{x}{n})^n$, where $n$ is an integer, then
$n$ must be chosen to find $\frac{x}{n}$ as close as possible to e.
Or in technical notation, to minimise the absolute value of
$\frac{x}{n}-e$ subject to $n$ an integer for given $x$.</p>
<p class="editorial">Well done Thomas, you've really got the hang
of this problem. For those who don't understand his notation, you
have to find an integer $n$ to make $\frac{x}{n}$ as close as
possible to 2.7.You then use this integer to find the answer.
Abover $x=10$, $n=4$ and so $\frac{x}{n}$ is 2.5. So the
sum is $10$, and the product is $(\frac{x}{n})^n$, which
is $39.0625$ in this case.</p>
<p class="editorial">Well done everyone on a very tough
problem!</p>
<br></br></mdoxml>
<mdoxml version="1.0"><br></br>
<h3>Why do this problem?</h3>
<div><a href="http://nrich.maths.org/public/viewer.php?obj_id=1785">This
problem</a> can be tackled at very different intellectual levels.
It is an excellent context to practise adding and multiplying with
decimals while trying to solve a problem.</div>
<br></br>
Part of the problem is to realise that there is, in fact, a
question to consider.<br></br>
<br></br>
<h3>Possible approach</h3>
<div>Ask the students for various sets of numbers that add to 10
(exclude negatives!). Ask students to work out the products of some
of these sets - students might demonstrate the methods on the board
to help the group remember the algorithms.</div>
<br></br>
<div>Present the problem for the lesson - to find the biggest
product, for numbers that add to $10$, to keep a sensible record of
things that they have tried and what did and didn't work out.</div>
<br></br>
<div>After some time working, students could feed back on
successful strategies they are using, and the class 'best' could be
put up on the board for others to beat. Make sure that students
know that there are ways of adapting the problem if they feel that
they are making no further progress.</div>
<h3>Key questions</h3>
<ul>
<li>
<p>How shall we start to make some progress with this challenge
?</p>
</li>
<li>
<p>Can you generalise your results somehow?</p>
</li>
<li>
<p>What is the same and what is different
about the solution if the $10$ is changed to another number?</p>
</li>
</ul>
<br></br>
<h3>Possible extension</h3>
<div>What about the sum being $0.5$ or $1.1$ or $-8$ etc., how
would the strategy/solution change?</div>
<br></br>
<div>Students often work by reasoning from a few 'spot' values and
showing that improvement occurs as they adjust towards a particular
value, which they therefore declare to be the optimum. It is
certainly good to draw attention to the logical possibility that an
even better 'local maximum' might exist somewhere else. And also,
if they are up to it, that adjustments are always incremental so
how can we be sure that we haven't jumped right over an interval
which contains something important.</div>
<div>Can we reason that this function is continuous ?</div>
<br></br>
<h3>Possible support</h3>
<div>With students who aren't ready for working with these
decimals, perhaps consider integer solutions only for various
sums.</div>
<br></br>
<br></br>
<br></br>
<br></br></mdoxml>
<mdoxml version="1.0">Thinking about rectangles and cuboids might be useful.<br></br>
Try some numbers out for yourself.<br></br></mdoxml>
<mdoxml version="1.0"><br></br>
<br></br>
<br></br>
the n value that maximises $(\frac{k}{n})^n$ is $k/e$ though
interestingly, this cannot be obtained by calculus (says Mike)
<br></br>
this is clearly beyond the scope of most stage three students (non
integer powers)though they may well graph the function
$(\frac{k}{n})^n$ for some k, or on autograph with varying k<br></br>
<br></br>
<br></br>
these are the student solutions from the original publication of
the problem :<br></br>
<br></br>
<p>When $10$ is divided into $2$ parts, the largest product is $5
\times5 = 25$.</p>
<br></br>
If it is divided into $3$ parts, the largest product is $3
{1\over3} \times3 {1\over3} \times3 {1\over3}$.<br></br>
<br></br>
If it is divided into $4$ parts, the largest product is $2
{1\over2}\times 2 {1\over2} \times2 {1\over2} \times2
{1\over2}$.<br></br>
<br></br>
If it is divided into $5$ parts, the largest product is $2\times 2
\times2 \times2 \times2$.<br></br>
<br></br>
It soon becomes apparent that when $10$ is split into $n$ parts
(where n is a whole number), the largest product is given by
$\left({10\over n}\right)^n$.<br></br>
<br></br>
So the problem is in fact to find the largest value of
$\left({10\over{n}}\right)^n$. <br></br>
<br></br>
The largest value of $\left({10\over{n}}\right)^n$ is when
$n=4$:<br></br>
<br></br>
$\left({10\over4}\right)^4 = 2.5^4 = 39.0625$ <br></br>
<br></br>
This is the largest product.<br></br>
<br></br>
Here is an explanation to why this works: if there are two numbers
added to make $10$ they can be thought of as making a rectangle and
by plotting length against height for different rectangles on a
graph it is possible to see that the maximum area occurs (the
product) when the length equals the breadth.<br></br>
<br></br>
The same type of argument can be made for three numbers forming a
cuboid. The greatest volume will occur when the three dimensions of
the cuboid are equal (a cube).<br></br>
<br></br>
So we can definitely show that this is the case for sums made from
two and three numbers.
<p>Correct solutions to this were sent in by:</p>
<table border="0">
<tbody>
<tr align="left" valign="top">
<td align="left" valign="middle"><strong>Hannah</strong></td>
<td align="left" valign="middle">Archbishop Sancroft High
School</td>
</tr>
<tr align="left" valign="top">
<td align="left" valign="middle"><strong>Stephen</strong></td>
<td align="left" valign="middle">Aylsham Middle School</td>
</tr>
<tr align="left" valign="top">
<td></td>
</tr>
<tr align="left" valign="top">
<td align="left" valign="middle"></td>
<td align="left" valign="middle"></td>
</tr>
</tbody>
</table>
<br></br>
<br></br></mdoxml>
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Largest product
Which set of numbers that add to 10 have the largest product?
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