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

Mar 10, 2016

$\left(\frac{101}{2} , \frac{101}{2}\right)$

#### 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

$y = x \cdot \left(101 - x\right) = - {x}^{2} + 101 x$
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.

$- {x}^{2} + 101 x = - \left({x}^{2} - 101 x\right)$

$= - \left({x}^{2} - 101 x + {\left(\frac{101}{2}\right)}^{2} - {\left(\frac{101}{2}\right)}^{2}\right)$

$= - {\left(x - \frac{101}{2}\right)}^{2} + {\left(\frac{101}{2}\right)}^{2}$

Given the vertex form $a {\left(x - h\right)}^{2} + k$ the vertex occurs at $\left(h , k\right)$, and thus the vertex in this case occurs at $\left(\frac{101}{2} , {\left(\frac{101}{2}\right)}^{2}\right)$

From this, we know that $x = 101 - x = \frac{101}{2}$, meaning the maximum product occurs when the pair of numbers is $\left(\frac{101}{2} , \frac{101}{2}\right)$.

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 $x = 101 - x$.