What is the smallest value of $x^2+y^2$ when $x+y=6$? If $ x+y=6 $ then what is the smallest possible value for $x^2+y^2$?
Please show me the working to show where I am going wrong!
Cheers
 A: You can solve this using geometry. $x+y=6$ is the equation of a line in the 2D plane. $x^2 + y^2$ is the squared distance to the origin. So you need to find the point on the line which is closest to the origin. This is obtained by orthogonally projecting the origin on the line, along the $x=y$ line.
Solving $x+y=6$ and $x=y$ gives the result $x=y=3$ and $x^2+y^2=18$.
A: $$x^2+y^2\geq 2xy\implies 2(x^2+y^2)\geq (x+y)^2$$
Hence, $$(x^2+y^2)\geq \frac{6^2}{2}=18$$
A: Reorder $x+y = 6$ to
$y=6-x$
Substituting $y$ in $(x^2+y^2)$ yields
$x^2+(6-x)^2 = 2x^2-12x+36$
The minimum occurs where the derivative equals $0$
$4x -12 = 0$
Therefore at the minimum,
$x=3$
Hence the minimum is $2\cdot 3^2-12\cdot 3+36 = 18$
A: Hint: Since $x + y = 6$, we find that
$$x^2 + y^2 = x^2 + (6 - x)^2 = 2x^2 - 12x + 36$$
This can be minimized in any number of ways, e.g. vertex formula or differentiating. 
A: We have 
$$2(x^2+y^2)=(x+y)^2 +(x-y)^2=36+(x-y)^2.$$ 
But $36+(x-y)^2$ is smallest when $x=y$. Thus the minimum value of $2(x^2+y^2)$ is $36$. 
A: Using Am-Qm inequality :
$\sqrt{(x^2+y^2)/2}>=(x+y)/2$
Solving this we get :
$x^2+y^2>=18$
For more on Am-Qm visit this site
Cheers
A: When Usually found yourself stuck at such questions:
One can Go for Hit and Trial Method:
Since the value of 
 x+y = 6
Start with lowest possible combination you can think of
For x=1, y=5 : $x^2+y^2=26$
For x=2, y=4 : $x^2+y^2=20$
For x=3, y=3 : $x^2+y^2=18$
For x=4, y=2 : $x^2+y^2=20$
For x=5, y=1 : $x^2+y^2=26$
So the minimum is 18
A: Most reasonable and concrete solution is to find minima using derivatives.
As shown by DeltaLima. 
There are many ways to find the minima graphically: 
$x^2+y^2$ expression can be written as $x^2+y^2=K$(equation)(it represents parabola) and $x+y=6$ represents, a straight line. They will intersect for minimum and maximum value of $(x,y)$.
A: If you believed $x=y=3$ is probably the answer, then you could shift the problem to be centered there: that often makes things clearer.
So define $x' = x - 3$ and $y' = y-3$. Our problem is now that $x' + y' = 0$ and we want to minimize $(x'+3)^2 + (y'+3)^2$. If we expand that, we get
$$ x'^2 + 6x' + 9 + y'^2 + 6y' + 9 = x'^2 + y'+2 + 6(x' + y') + 18 = x'^2 + y'^2 + 18$$
This is obviously minimized when $x' = y' = 0$, as desired.
The above would look essentially the same if you solved $x+y=6$ for $y$ to rewrite the entire problem in terms of $x$ before doing the above.
