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I am not sure if my method for this question is correct: Given the function $$f(x,y)=x^2+y^2+2x+y$$, find it's minimum and maximum values about a closed disc of radius 2 centred at the origin. I took the partial derivatives, $$f_x=2x+2; f_y=2y+1$$ and solved for $x$ and $y$. The minimum value is $f(0,0)$ and the maximum value is $f(-1,-1/2)$. Am I doing the correct thing? The answer I got is wrong according to the key.

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If you have a boundary which is given by the level curve of a function (here $x^{2} + y^{2}$), you can use the Lagrange multiplier theorem (see here). You have to check the points of the boundary such that the gradient of the function you want to study and the gradient of the function giving the domain are parallel.

In this case you have that the only critical point in $\mathbb{R}^{2}$ is $\left(-1 , -1/2 \right)$. $f$ is the sum of two quadratic functions in $x$ adn $y$ separately and with coefficients of $x^{2}$ and $y^{2}$ both positive. Hence $\mathrm{lim}_{||\left(x,y\right)|| \to +\infty}f\left(x,y\right)=+\infty$, so the point we have is the absolute minimum, and it lies in the circle. Now to find the maximum in the circle check the points of the circle of radius 2 such that $\left(2x+2,2y+1\right)$ is parallel to $\left(2x,2y\right)$. If you put them in a matrix and want the determinant to be $0$, it gives you the condition $2y=x$.

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  • $\begingroup$ One question. Do I substitute $2y=x$ into $x^2+y^2-4=0$? $\endgroup$
    – Artemisia
    Nov 19 '13 at 10:30
  • $\begingroup$ Yes, you should find the intersection between the line $2y=x$, which gives you the condition for the Lagrange multipliers, and the locus you have, i.e. the circle $\endgroup$
    – Stefano
    Nov 19 '13 at 10:32
  • $\begingroup$ So I get $y=\sqrt{4/5}$ and $x=\sqrt{8/5}$ which gives function value of $7.20$. This is incorrect according to the key. $\endgroup$
    – Artemisia
    Nov 19 '13 at 10:34
  • $\begingroup$ The line intersects the circle in two points, try with a minus... $y=\pm \sqrt{4/5}$ and $x=\pm \sqrt{16/5}$. $\endgroup$
    – Stefano
    Nov 19 '13 at 10:38
  • $\begingroup$ But if it asks for maximum value... then won't addition give a higher value? The answer I get is $-0.8$ $\endgroup$
    – Artemisia
    Nov 19 '13 at 10:41
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I'm not sure where you got the $(0,0)$ from. It is not a critical point--and you should find for example that $f(0,-\frac12)<f(0,0)$. Your second point is the only critical point, and it lies in the given disc.

It remains to check the boundary of the disc--that is, the points $(x,y)$ such that $x^2+y^2=2^2$--for critical points. I suggest you use the substitutions $x=2\cos\theta$ and $y=2\sin\theta,$ to get a function $g(\theta)=f\bigl(x(\theta),y(\theta)\bigr).$ Then you can use that function to find the critical points on the boundary (if any). Once you've found all critical points on the interior and boundary, you can check to see what the maximum and minimum values of $f$ are.

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  • $\begingroup$ Oh so in order to obtain these values, I need to check the critical point and the boundary? I graphed the function and saw its behavior at the origin so I thought it could be considered. Also, the function expands into infinity. Does it have a boundary at all? $\endgroup$
    – Artemisia
    Nov 18 '13 at 16:42
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    $\begingroup$ I'm not sure what "behavior at the origin" you're talking about. We definitely need to check the boundary, too, though. Extreme values will occur at critical points on the interior of the region, or else they will occur on the boundary. $\endgroup$ Nov 18 '13 at 16:44
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By "completing the squares" for the terms in each coordinate variable, we obtain

$$ f(x,y) \ = \ x^2 \ + \ y^2 \ + \ 2x \ + \ y \ \ = \ \ (x \ + \ 1)^2 \ - \ 1 \ + \ \left(y \ + \ \frac{1}{2} \right)^2 \ - \ \frac{1}{4} $$ $$= \ \ (x \ + \ 1)^2 \ + \ \left(y \ + \ \frac{1}{2} \right)^2 \ - \ \frac{5}{4} \ \ , $$

so there is indeed a critical point at $ \ \left( -1 \ , \ -\frac{1}{2} \right) \ $ where the value of the function is $ \ -\frac{5}{4} \ \ . $ This is the only critical point in the interior of the disk; since we see immediately that $ \ f(0 \ , \ 0) \ = \ 0 \ \ , $ this is a local minimum for the function. [We can confirm this through the usual multivariate test: $ \ f_{xx}f_{yy} - [f_{xy}]^2 \ = \ 2 · 2 \ - \ 0^2 \ > \ 0 \ $ and $ \ f_{xx} \ > \ 0 \ . $ ] There is nothing special about the function at the origin.

Since the region of interest is the closed disk $ \ x^2 \ + \ y^2 \ \le \ 4 \ , \ $ we will also want to explore the boundary. There are a number of ways to do so: here, we'll look at a couple not already discussed in the other postings. As we are working on the boundary $ \ x^2 \ + \ y^2 \ = \ 4 \ , \ $ we could write the function as $ \ f(x,y) \ = \ 4 \ + \ 2x \ + \ y \ , \ $ and focus only on the last two terms.

If we use Cartesian coordinates, we can form the single-variable function $ \ \phi(x) \ = \ 2x \ + \ \sqrt{4 \ - \ x^2} \ $ , for which the extremization gives us

$$ \ \phi'(x) \ = \ 2 \ - \ \frac{x}{\sqrt{4 \ - \ x^2}} \ = \ 0 \ \ \ \Rightarrow \ \ \ 16 \ - \ 4x^2 \ = \ x^2 \ \ \ \Rightarrow \ \ x \ = \ \pm \frac{4}{\sqrt{5}} $$ $$ \Rightarrow \ \ \ y \ = \ \pm \frac{2}{\sqrt{5}} \ \ . $$ The absolute maximum on the disk is then $$ \ f\left(\frac{4}{\sqrt{5}} \ , \ \frac{2}{\sqrt{5}} \right) \ = \ 4 \ + \ 2·\frac{4}{\sqrt{5}} \ + \ \frac{2}{\sqrt{5}} \ = \ 4 \ + \ \frac{10}{\sqrt{5}} \ = \ 4 \ + \ 2\sqrt{5} \ \approx \ 8.472 $$

and the local minimum on the boundary is $ \ f\left(-\frac{4}{\sqrt{5}} \ , \ -\frac{2}{\sqrt{5}} \right) \ = \ 4 \ - \ 2\sqrt{5} \ \approx \ -0.472 \ \ . $ So in fact the critical point at $ \ \left( -1 \ , \ -\frac{1}{2} \right) \ $ is the location of the absolute minimum value $ \ -\frac{5}{4} \ $ for the region.

We could also find the extrema on the boundary without calculus, by employing a method we learn for working with trigonometric functions. Using polar coordinates as suggested by Cameron Buie, we can write our one-variable function as $ \ \phi(\theta) \ = \ 2 · 2 \cos \theta \ + \ 2 \sin \theta \ \ . $ Extracting a factor of $ \ \sqrt{4^2 \ + \ 2^2} \ $ produces

$$ \ \phi(\theta) \ = \ \sqrt{20} \ \left[ \ \underbrace{\frac{4}{\sqrt{20}}}_{\cos \alpha} \ \cos \theta \ + \ \underbrace{\frac{2}{\sqrt{20}}}_{\sin \alpha} \ \sin \theta \ \right] \ \ = \ \ 2\sqrt{5} \ \cos (\theta \ - \ \alpha) \ \ , $$

making use of an "angle-addition" formula, where $ \ \alpha \ $ has the indicated trigonometric values. We see directly that $ \ \phi(\theta) \ $ has the extremal values $ \ \pm \ 2\sqrt{5} \ $ , so the corresponding values of our function $ \ f(x) \ $ on the boundary are $ \ 4 \ \pm \ 2\sqrt{5} \ \ . $

(While it is not required for the given problem, should we wish to know the locations of the extremal points, they are at a radius of $ \ 2 \ $ from the origin at the angles corresponding to

$$ \tan \alpha \ \ = \ \ \frac{\frac{2}{\sqrt{20}}}{\frac{4}{\sqrt{20}}} \ \ = \ \ \frac{1}{2} \ \ , $$

with the maximum at $ \ \theta \ = \ \alpha \ $ and the minimum at $ \ \theta \ = \ \alpha \ + \ \pi \ \ . $ This agrees with the rectangular coordinates we found earlier.)

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