# Thread: Two numbers have a sum of 13 (terminology & help)

1. ## Two numbers have a sum of 13 (terminology & help)

1. The problem statement, all variables and given/known data
10) Two numbers have a sum of 13.
10)a) Find the minimum of the sum of their squares.
10)b) What are the two numbers

2. Relevant equations
y = ax^2 + bx + c
y = a (x-h)^2 + k

According to the text: for a quadratic function in the forum of y=a(x-h)2+k, the maximum or minimum value is k, when x=h. If a> 0, k is the minimum value of the function. If a <0, k is the maximum value of the function.

3. The attempt at a solution

No attempt, do not understand how to properly attempt the question.

What I believe to understand is that
a) the question is asking for the value of two numbers which add up to 13, and
b) what the value of those two numbers squared, then added up together is. I don't understand why it's asking for the "minimum" value of their squares.

2. Originally Posted by Monocerotis
1. The problem statement, all variables and given/known data
10) Two numbers have a sum of 13.
10)a) Find the minimum of the sum of their squares.
10)b) What are the two numbers
let $x$ = 0ne of the numbers

$13-x$ = the other number

sum of their squares ...

$x^2 + (13-x)^2$

expand the quadratic expression and locate the vertex, which will be the location for the minimum ... then determine the two numbers.

3. Originally Posted by Monocerotis
1. The problem statement, all variables and given/known data
10) Two numbers have a sum of 13.
10)a) Find the minimum of the sum of their squares.
10)b) What are the two numbers
I think you'll have to find two numbers a and b such as a+b=13 and $a^2+b^2$ has the smallest possible value. I think the minimum is obtained when a=b, so a=b=6.5

4. Originally Posted by skeeter
let $x$ = 0ne of the numbers

$13-x$ = the other number

sum of their squares ...

$x^2 + (13-x)^2$

expand the quadratic expression and locate the vertex, which will be the location for the minimum ... then determine the two numbers.
We just started learning about the quadratic function last class, how do you "expand the quadractic expression and locate the vertex"

$x^2 + (13-x)^2$ looks a bit like $y= a(x-h)^2+k$, but other than noticing a minor similarity I'm just lost. :S

5. Before getting all excited, you could do a li'l look-see:

0,13 : 0 + 169 = 169
1,12 : 1 + 144 = 145
2,11 : 4 + 121 = 125
...
6,7 : 36 + 49 = 85
6.5,6.5 : 42.25 + 42.25 = 84.5

Now at least you know what "you're aiming for"

6. Originally Posted by Wilmer
Before getting all excited, you could do a li'l look-see:

0,13 : 0 + 169 = 169
1,12 : 1 + 144 = 145
2,11 : 4 + 121 = 125
...
6,7 : 36 + 49 = 85
6.5,6.5 : 42.25 + 42.25 = 84.5

Now at least you know what "you're aiming for"
Yeah I know the answer is 6.5, 6.5

It's simple enough to just divide 13 by 2 and then square those two numbers, logically you would arrive with the sum being the smallest of all possible numbers.

The problem is working out the correct way to do it, involving the quadratic formula.

I'm taking this math course at night school, which means the program is accelerated and the teacher is less willing to help, for both person reasons (bad teacher), and the fact that they are probably tired from teaching all day already.

So I am exercising the option to ask the internet (math help forum).

7. Just another way using quadratic:

The 2 numbers are n, 13-n ; m = minimum

n^2 + 169 - 26n + n^2 = m
2n^2 - 26n + 169 - m = 0
1st derivative:
4n - 26 = 0
4n = 26
n = 6.5 ; 13 - 6.5 = 6.5