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Find the maximum and minimum values of $f(x,y)=xy-2x$ on the rectangle $-1\leq x \leq1 $ and $0\leq y \leq 1$.

I don't understand the approach. The solution manual suggests that the critical point is not inside my domain so maximum and minimum values of $f$ must be on one of the four boundary points. I don't understand how we get to this conclusion.

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  • $\begingroup$ write $f(x,y)=x(y-2)$ and note $y-2<0$. So, intuitively, $f(-1,0)=2$ (max) and $f(1,0)=-2$ (min). $\endgroup$ – farruhota May 2 '19 at 12:24
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If you take the gradient of $f$ you have $$ \nabla f = ( y - 2, x)' = (0,0)', $$ i.e., the critical point is $(0,2)$ which lies outside of your domain. As such, you'll get the global maximum and minimum in $D$ on $\partial D$. Namely, check for $-1 \le x \le 1$ and $y=0$, and you'll get $$ g(x)= -2x, $$ that is monotone decreasing function, i.e., $(-1, 0)$ is a candidate point. For $y=1$ you have $$ g(x) = x - 2x, $$ hence $(-1, 1)$ is a candidate point. Then, check $x=1$ and $0 \le y \le 1$, thus $$ g(y)= y - 2, $$ i.e., $(1,1)$i is another candidate. For $x=-1$ you'll get $$ g(y)=-y+2, $$
so $(-1,0)$ is another point. Finally, check the connection points of the $4$ parts of $\partial D$, i.e., $(-1, 0)$, $(-1,1)$, $(1, 0)$ and $(1,1)$ that are mostly the same candidates as in the aforementioned analysis. Finally, just compare $f$ in all of these points.

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Check where the derivatives are zero in the domain given, if it's not zero anywhere in the domain then the maximum or the minimum would occur at the extreme values.

This will help http://personal.maths.surrey.ac.uk/st/S.Zelik/teach/calculus/max_min_2var.pdf

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