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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

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11  
Why not show what you've tried? That way we can tell you where you went wrong. –  Michael Albanese Sep 23 '13 at 3:49
4  
">Please show me the working to show where I am going wrong!" You haven't provided any work for us to check –  Ozera Sep 23 '13 at 5:12

9 Answers 9

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 $36$.

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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.

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4  
To add another familiar phrase to the list: "completing the square". –  Hurkyl Sep 23 '13 at 10:48

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 $x^2+y^2=18$.

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$$x^2+y^2\geq 2xy\implies 2(x^2+y^2)\geq (x+y)^2$$

Hence, $$(x^2+y^2)\geq \frac{6^2}{2}=18$$

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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 $2\cdot 3^2-12\cdot 3+36 = 18$

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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

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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 $(x,y)$.

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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

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3  
I don't think that suggesting trial and error here is a good idea. There's no reason to believe that the minimum will be when $x$ and $y$ are both integers. –  user61527 Sep 23 '13 at 7:29
    
or u can just say when all the quantities are equal ..either that will be the minimum or maximum as per the question –  maths lover Sep 23 '13 at 7:42

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.

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