Extending Multiplication to Algebra and Calculus

In my earlier post on a multiplication trick for young learners, I forgot to include a higher level connection, so I’ll do that here.

LEVEL 6:  Advanced Algebra or Calculus.
Part of what makes this problem work out so nicely is that the sums of the numbers involved in the lists are constant, an omnipresent property of position-symmetric terms from any arithmetic sequence.  That is, given any arithmetic sequence of sufficient length (e.g., {1, 2, 3, 4, 5, 6, 7, 8}), the first and last terms (1 & 8) will always have the same sum as the terms just inside those (2 & 7), the next terms inside (3 & 6), and so forth until you run out of terms.  This property can be proved for any arithmetic sequence with introductory algebra.

Here’s the advanced math connection.  Given all pairs of numbers which add to some constant sum, k, what is the relationship between the pair of these numbers whose product is maximum?

Let the two numbers be X and Y where X+Y=k.  We want to find the maximum of the product X\cdot Y=X\cdot (k-X).  If you recognize this last expression as a quadratic function, f(X)=X\cdot (k-X), then you can use the location of the vertex of a parabola (algebra) or some calculus (find the zero of \displaystyle\frac{df}{dX}) to see that the maximum of f occurs at \displaystyle X=\frac{k}{2}=Y.

Complicated math aside, this means that when you have a list of pairs of numbers that all add to the same constant, the pair with the biggest product is always the pair of numbers closest in proximity to each other.  In the context of this problem, this explains why the inner products of any arithmetic sequence are always larger than the products of the first and last terms.

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