Finding x in quadratic function

Question

The sum of two numbers is 10. If the sum of their squares is a minimum, what are the two numbers? I can't think of how to write an equation to solve this question as I dont know what are the coordinates of the vertex. Can anyone please teach me? Thank you!

Answer 1

Let the two numbers be $5-x$ and $5+x$

The sum of their squares is

$(5-x)^2 + (5+x)^2 = 50 + 2x^2$

which clearly becomes least when $x=0$ Hence the numbers are both 5

Answer 2

Let the numbers be $a \le b$ with $b \ge 0$ since $a+b =10 \ge 0$. We can assume that $a \ge 0$ otherwise the pair $0, b+a$ would have the same sum, but the sum of squares would go lower since $(b+a)^2 =a^2 + b^2 + 2 ab\le a^2 + b^2$ for $a \le 0 \le b$.

Then by the root-mean square inequality:

$$\sqrt{\frac{a^2+b^2}{2}} \ge \frac{a+b}{2} = \frac{10}{2}=5$$

with equality iff $a=b$. Therefore the minimum of the $LHS$ is achieved when $a=b=\frac{a+b}{2}=5$.

Answer 3

Letting one number be $x$, the other is $10-x$ and the sum of their squares is $x^2 + 100 + x^2 - 20x = 2(x^2 - 10x + 50) = 2((x-5)^2 + 25)$ (the last step involves completing the square).

Since $(x-5)^2$ is always non-negative for real $x$, its minimum is zero and that occurs when $x = 5$, so the two numbers are $5$ and $5$ and the minimum sum of their squares is $50$.

Answer 4

Just another way.

Call your number $x$ and $y$. You know that $x+y=10$ and you want

$$F=x^2+y^2$$ to be minimum. From the containt, $y=10-x$; replace in $F$

$$F=x^2+(10-x)^2=x^2+(100-20x+x^2)=2x^2-20x+100$$ Since it is a minimum, the derivative must be zero. Computing $$F'=(2x^2-20x+100)'=4x-20=0\implies x=5\implies y=5$$