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Find the maximum and minimum of $f(x, y)=xy-y+x-1$ at the set $x^2+y^2\leq 2$.

I have done the following:

Since the region $x^2+y^2\leq 2$ is closed, $f$ has a maximum and a minimum, which is either at the boundary or at the critical points of the function.

To find the critical points we do the following:

$$\nabla f=\overrightarrow{0} \Leftrightarrow (y+1, x-1)=(0, 0) \Rightarrow x=1, y=-1$$

So, we have that the only critical point is $(1, -1)$, which satisfies the contraint.

To find the extremas of $f$ at the boundary $x^2+y^2=2$, we will use Lagrange multipliers therem.

The constraint is $g(x, y)=x^2+y^2-2=0$.

We are looking for $x$, $y$ and $\lambda$ such that $$\nabla f(x, y)=\lambda \nabla g(x, y) \tag 1$$ and $$g(x, y)=0 \tag 2$$

$$(1) \Rightarrow (y+1, x-1)=\lambda (2x, 2y) \Rightarrow \lambda =\frac{1}{2}$$ $$\Rightarrow x=y+1$$

$$(2) \Rightarrow y=\frac{-1+\sqrt{3}}{2}, y=\frac{-1-\sqrt{3}}{2}$$

SO, the extremas are the points $\left (\frac{1+\sqrt{3}}{2}, \frac{-1+\sqrt{3}}{2}\right )$ and $\left (\frac{1-\sqrt{3}}{2}), \frac{-1-\sqrt{3}}{2}\right )$.

Calculating the values of $f$, we have: $$f\left (\frac{1+\sqrt{3}}{2}, \frac{-1+\sqrt{3}}{2}\right )=\frac{1}{2} \\ f\left (\frac{1-\sqrt{3}}{2}), \frac{-1-\sqrt{3}}{2}\right )=\frac{1}{2} \\ f(1, -1)=0$$

SO, $f$ has a local maximum at the points $\left (\frac{1+\sqrt{3}}{2}, \frac{-1+\sqrt{3}}{2}\right )$ and $\left (\frac{1-\sqrt{3}}{2}), \frac{-1-\sqrt{3}}{2}\right )$ and an absolute minimum at $(1, -1)$.

Is this correct?? Could I improve something??

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If you wanted to check the answer you get from the Lagrange multipliers, you could look for "extrema" along the boundary of the region by looking at the map:

$$F: \mathbb R \rightarrow \mathbb R^{2} \rightarrow \mathbb R$$ $$F(t) = f(\gamma(t)) = f(\sqrt{2} \cdot cos(t), \sqrt{2} \cdot sin(t)), t \in [0, 2\pi]$$

Which is a function on one variable relating every point in the boundary to its corresponding "height" in the graph of $z = f(x, y)$.

The resulting function is:

$$F(t) = f(\gamma(t)) = 2 \cdot cos(t) \cdot sin(t) - \sqrt{2} \cdot sin(t) + \sqrt{2} \cdot cos(t) - 1$$

From here you may find its derivative, etc. with all the tools from functions in one variable.

(To get $\gamma(t)$ you just parametrize the boundary of the region using the identity $cos(t) + sin(t) = 1$.)

I hope this helps, I find it useful to be able to check this sort of problem with a "1D version", so to speak.

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  • $\begingroup$ PS.: for a quick "visual check" you could just plot this function in WolframAlpha, etc. without having to resort to any "rigorous" analysis $\endgroup$ – shost71 May 29 '15 at 0:32
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Looks good, I have just a few comments:

"So the extras are the points..." at this point we don't know which of these is an extremum. These are possible extreme points. It is only after we plug in all the values that we know which are extrema.

"$f$ has a maximum and minimum" I would phrase as "$f$ achieves its maximum and minimum"

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