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

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

$S = {n}^{2} + {\left(n - 20\right)}^{2} = 2 {n}^{2} - 40 n + 400$

As we can see, $S \left(n\right)$ is a parabola with a minimum at

$\frac{d}{\mathrm{dn}} S \left({n}_{0}\right) = 4 {n}_{0} - 40 = 0$ 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 $\frac{d}{\mathrm{dn}} S \left({n}_{0}\right)$ is Calculus. Let’s see if we can solve this without calculus.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{m a \ge n t a}{\text{Let the first number be } x}$
Let the second number be $x + 20$

Set $\text{ } y = {x}^{2} + {\left(x + 20\right)}^{2}$

$y = {x}^{2} + {x}^{2} + 40 x + 400$

$y = 2 {x}^{2} + 40 x + 400 \leftarrow \text{ "y" is the sum of their squares}$

$\textcolor{red}{\text{So we need to find the value of x that gives the minimum value}}$ $\textcolor{red}{\text{of } y}$

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

Write as $y = 2 \left({x}^{2} + 20 x\right) + 400$

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

Consider the 20 from $20 x$

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

Thus the first number is $x = - 10$
The second number is $\text{ } x + 20 = - 10 + 20 = 10$

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