Two numbers have a difference of 20. How do you find the numbers if the sum of their squares is a minimum?

2 Answers
Dec 9, 2016

-10,10

Explanation:

Two numbers n,m such that n-m=20

The sum of their squares is given by

S=n^2+m^2 but m = n-20 so

S=n^2+(n-20)^2=2n^2-40n+400

As we can see, S(n) is a parabola with a minimum at

d/(dn)S(n_0) = 4n_0-40=0 or at n_0 = 10

The numbers are

n=10, m=n-20=-10

Dec 9, 2016

10 and -10

Solved without Calculus.

Explanation:

In Cesareo’s answer d/(dn)S(n_0) is Calculus. Let’s see if we can solve this without calculus.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(magenta)("Let the first number be "x)
Let the second number be x+20

Set " "y=x^2+(x+20)^2

y=x^2+x^2+40x+400

y=2x^2+40x+400 larr" "y" is the sum of their squares"

color(red)("So we need to find the value of x that gives the minimum value") color(red)("of "y)

This equation is a quadratic and as the x^2 term is positive then it its general shape is of form uu. Thus the vertex is the minimum value for y

Write as y=2(x^2+20x)+400

What follows is part of the process for completing the square.

Consider the 20 from 20x

color(magenta)("Then the first number is: "x_("vertex")=(-1/2)xx20= -10)

Thus the first number is x=-10
The second number is " "x+20 =-10+20 =10

" "color(green)(bar(ul(|color(white)(2/2)"The two numbers are: -10 and 10 "|)))

Tony B