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

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#.