Question

Among all pairs of numbers with a sum of 101, how do you find the pairs whose product is maximum?

Answer 1

`(101/2,101/2)`

Explanation:

Suppose one of the numbers is `x`. Then we know the other number is `101-x` and we wish to maximize their product which we call `y`, where

`y = x * (101-x) = -x^2+101x`
graph{x(101-x) [-100, 200, -5896, 4340]}
As this is a downward facing parabola, its maximum will occur at its vertex. Then, we can find the maximum by putting the quadratic into vertex form. We will do so using a process called completing the square.

`-x^2+101x = -(x^2-101x)`

`=-(x^2-101x+(101/2)^2-(101/2)^2)`

`=-(x-101/2)^2+(101/2)^2`

Given the vertex form `a(x-h)^2+k` the vertex occurs at `(h,k)`, and thus the vertex in this case occurs at `(101/2, (101/2)^2)`

From this, we know that `x = 101-x = 101/2`, meaning the maximum product occurs when the pair of numbers is `(101/2, 101/2)`.

Intuitively this also makes sense, as the problem could also be interpreted as asking what side lengths create the rectangle with the greatest area given a fixed perimeter of `202`. As a square has the maximal area among rectangles given a fixed perimeter, the solution should occur when all sides have equal length, that is, when `x = 101-x`.