# 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?

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.

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 $$50$$.

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 $$B: x = y = 5$$.

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.

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 $$(5,5)$$.

$$\large \text{How about this approach:}$$

$$x + y = 10, \text{ so } y = 10 - x \\ x^2 + y^2 = k, y = \sqrt{k - x^2}\\ \rightarrow 10 - x = \sqrt{k - x^2}\\ \rightarrow 100 - 20x + x^2 = k - x^2\\ \rightarrow 2x^2 - 20x + 100 = k\\ f\prime = 4x - 20\\ \text{When x is 5, } f\prime = 0\\~\\ \text{Testing values on the right and left: }\\ f(4) = 62, f(6) = 52\\~\\ \therefore \textbf{There is a minimum point at } x = 5, \textbf{ hence } y = 5.\\ 5^2 + 5^2 = 50$$

$$x^2+y^2\:=\: (x+y)^2-2xy \:=\: 100 -2xy$$ 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 $$\:\ddot\smile$$

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: $$\min (x^2+y^2)=5^2+5^2=50$$

• Thanks, I fixed it. – Max Feb 11 at 15:01

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 ($$x,y\geq0)$$: $$(x-y)^2\geq 0\iff x^2+y^2\geq 2xy\iff 2(x^2+y^2)\ge (x+y)^2\iff\color{red}{\sqrt{\frac{x^2+y^2}{2}}\geq \frac{x+y}{2}}$$