Max/min value in circle

Question

I'm working on this problem:

What is the max/min value of $x^2+y^2-4x-4y + 3$, in the space that is defined by $x^2+y^2<16$?

I have rewritten it as $x^2 + y^2-4x-4y+8-5$ which be expressed as $$(−2)^2+(−2)^2-5 .$$

But I don't know how to continue from there. How do I find the maximum and minimum value?

Answer 1

If you give an equation like $x^2+y^2=16$ for constrain you will find :

$x =y=\pm2\sqrt 2$, which gives $x^2+y^2=16$ and is acceptable.

As an standard method you must use Lagrange multiplier:

$f(x, y)=(x-2)^2+(y-2)^2-5$

$g(x, y)=x^2+y^2-16$

$\mathcal{L}_{x, y, \lambda}=f(x, y)+\lambda \cdot g(x,y)=(x-2)^2+(y-2)^2-5+\lambda(x^2+y^2-16)$

$\nabla_{x, y, \lambda}\mathcal{L}(x, y, \lambda)=\big(\frac{\delta\mathcal{L}}{\delta x},\frac{\delta\mathcal{L}}{\delta y},\frac{\delta\mathcal{L}}{\delta \lambda}\big)=(2x-4+2\lambda x),(2y-4+2\lambda y), (x^2+y^2-16)$

$\begin{cases}2x(1+\lambda)-4+0\\2y(1+\lambda)-4=0\\x^2+y^2-16=0\end {cases}$

which gives:

$x=y=\frac2{1+\lambda}$

putting this in third equation gives:

$\lambda=\frac{1-\sqrt2}{\sqrt 2}$ and $\lambda=-\frac{1+\sqrt 2}{\sqrt 2}$

which finally gives:

$x =y=\pm2\sqrt 2$

So minimum is :

$f(2\sqrt 2, 2\sqrt2)=24-16\sqrt2$

And maximum is :

$f(-2\sqrt 2, -2\sqrt 2)=24+16\sqrt 2$

Answer 2

I will look for the extremal values in the closed region $x^2+y^2\le 4^2$, as specified in the comments.

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For the maximal value, use $|x+y|^2=(x+y)^2\le 2(x^2+y^2)\le 32$ (with equality iff $x=y$) to get $$ x^2+y^2-4x-4y+3\le x^2+y^2+4|x+y|+3\le 16+4\sqrt{32}+3=19+16\sqrt 2\ . $$ The equality is obtained if we insure the equal sign at all $\le$-places where we have estimated above. So we need $x^2+y^2=16$, $x=y$, $-4(x+y)=4|x+y|$, and there is only one point, the one with $x=y=-2\sqrt 2$.

This agrees with wolfram alpha.

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On the other hand, forgetting about any restriction on $(x,y)$, we have - as in the question: $$ x^2+y^2-4x-4y+3= x^2+y^2-4x-4y+4+4-5= (x-2)^2+(y-2)^2-5\ge -5\ . $$ The equality is obtained in $(2,2)$, the absolute minimal value on $\Bbb R^2$ is obtained in this point, and yes, this point is in our domain.

This agrees with the other wolfram alpha request.