Minimum point of $x^2+y^2$ given that $x+y=10$ How do you I approach the following question:
Find the smallest possible value of $x^2 + y^2$ given that $x + y = 10$.
I can use my common sense and deduce that the minimum value is $5^2 + 5^2 = 50$. But how do you approach this mathematically?
Thanks in advance.
 A: $$x^2+y^2\:=\: (x+y)^2-2xy \:=\: 100 -2xy$$
is minimal if $\,xy\,$ is maximal (hence $x,y\geqslant 0$). Which is maximal if $\,\sqrt{xy}\:$ is maximal.
The inequality of arithmetic and geometric means
$$\sqrt{xy}\:\leqslant\:\frac{x+y}2 \:=\: 5$$
gets an equality only for $y=x$, which means the LHS is maximal.
Thus your common sense is mathematically, especially algebraically, confirmed $\:\ddot\smile$
A: $\large \text{How about this approach:}$
$
x + y = 10, \text{ so } y = 10 - x \\
x^2 + y^2 = k, y = \sqrt{k - x^2}\\
\rightarrow 10 - x = \sqrt{k - x^2}\\
\rightarrow 100 - 20x + x^2 = k - x^2\\
\rightarrow 2x^2 - 20x + 100 = k\\
f\prime = 4x - 20\\
\text{When x is 5, } f\prime = 0\\~\\
\text{Testing values on the right and left: }\\
f(4) = 62, f(6) = 52\\~\\
\therefore \textbf{There is a minimum point at } x = 5, \textbf{ hence } y = 5.\\
5^2 + 5^2 = 50
$
A: Since the condition $x+y=10$ is simple and allows easy elemination of one variable, one possible approach is to put the resulting $y=10-x$ into the term to mimimize:
$$x^2+y^2=x^2+(10-x)^2:=f(x)$$
You now have a function in one variable (named $f(x)$) where you are looking for the minimal value over all real $x$.
Since this is a quadratice function, it can be minimized with a bit of algebra and without calculus:
$$f(x)=x^2+(10-x)^2=2x^2-20x+100=2(x^2-10x)+100 = 2(x^2-10x+25) + 50 = 2(x-5)^2+50 \ge 50.$$
So we have $f(x)\ge 50$ and equality happens at $x=5$ (which implies $y=5$).
Since the tag "derivatives" is used, I'll also use the usual calculus approach:
We have $f(x) = 2x^2-20x+100$, which implies $f'(x)=4x-20$ and $f''(x)=4$. 
A local mimimum $x_m$ has $f'(x_m)=0$ as necessay condition, and $4x_m-20=0$ easily leads to the only solution $x_m=5$ and $f''(x_m)=4 > 0$ shows this is a local mimimum, with $f(x_m)=f(5)=50$.
Also, one needs to check the behaviour of $f(x)$ when $x$ tends to $+\infty$ and $-\infty$, as $f'(x_m)=0$ only finds local extrema. Since $f(x)$ is a quadratic with positive constant before $x^2$, the function tends to $+\infty$ in either case, so no interference with the looked for minimum.
A: Take the vectors $u=(1,1)$ and $v=(x,y)$ then apply Cauchy-Schwarz inequality. It comes : $2(x^2+y^2)\geq (x+y)^2=100$ and for $x=y=5$ the equality holds. The minimum value is therefore $50$.
A: Showing the problem graphically

The line is defined by the equation $f: x + y = 10$. Minimizing $x^2 + y^2$ amounts to finding the smallest circle around the origin touching the line. Since all radii are perpendicular, we search for the intersection of $f$ and $y - x = 0$ which is solved by $B: x = y = 5$.
A: The general, "If all you have is a hammer, everything looks like a nail" method requiring very little creative thinking is to use Lagrange multipliers. 
(Note that there are some nontrivial conditions on when the method of Lagrange multipliers can be used; for example things get a bit messier if $\nabla g(x,y) = 0$ is possible when $g(x,y)=0$.)
You want to minimize $f(x,y) = x^2 + y^2$ subject to the condition $g(x,y)=x+y-10 = 0$. The point of using Lagrange multipliers is that you get simple conditions for the critical point of the constrained problem with the cost of having to add another unknown, $\lambda$, to the problem:
$$
\begin{align*}
\frac{\partial}{\partial x} f(x,y) &= \lambda \frac{\partial}{\partial x} g(x,y), \\
\frac{\partial}{\partial y} f(x,y) &= \lambda \frac{\partial}{\partial y} g(x,y), \\
g(x,y) &= 0.
\end{align*}
$$
Plugging in $f$ and $g$ there gives
$$
\begin{align*}
2x &= \lambda, \\
2y &= \lambda, \\
x+y - 10 &= 0,
\end{align*}
$$
which is a system of linear equations for $3$ variables, giving you $x=y=5$.
This method might seem like an overkill for such a simple problem, but once you're familiar with it, it's quite straightforward and effortless to write down the equations.
A: You can approach it geometrically. 
Condition $x+y=10$ means that the solution lies on the line $y = -x + 10$. $x^2+y^2$ is the square of distance between $(0,0)$ and $(x, y)$. That means that you want the closest point on the line $y = -x + 10$ to the origin. To get this, project the origin onto the line to get $(5,5)$. 
A: 
To minimize $f(x,y) = x^2+y^2$; given that $y=10-x$.

Then:
$$f(x,10-x)=x^2+(x-10)^2$$
$f$ is minimized for $f'=0$, so:
$$f'=2x+2(x-10)=0$$
Gives:
$$x=5\land y=5$$
Thus we have:
$$\min (x^2+y^2)=5^2+5^2=50$$
A: Most answers use functions or derivatives... I'll use another approach: Inequalities!
It's easy to prove that the minimum will be achieved for positive values of $x$ and $y$. Thus, in virtue of the QM-AM inequality: 

$$\sqrt{\frac{x^2+y^2}{2}}\geq \frac{x+y}{2}=5\iff x^2+y^2\geq 50$$


Proof of the Quadratic Mean - Arithmetic Mean inequality ($x,y\geq0)$: $$(x-y)^2\geq 0\iff x^2+y^2\geq 2xy\iff 2(x^2+y^2)\ge (x+y)^2\iff\color{red}{\sqrt{\frac{x^2+y^2}{2}}\geq \frac{x+y}{2}}$$
