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Find the absolute minimum and absolute maximum of the function f(x,y)=xy−1y−1x+1 on the region on or above y=x^2 and on or below y=4 and list the points where they occur.

So when doing this question I got f(1, 1) = 0 on the interior. However, I am confused with how I might find the boundary points when dealing with a question like this.

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Along the boundary $y=4$,

$$f(x,4)=-3+3x$$

which has the minimum at $f(-2,4) = -9$ and the maximum $f(2,4) = 3$.

Along the boundary $y=x^2$,

$$g(x)=f(x,x^2)=x^3-x^2-x+1$$

Setting $g'(x) = 0$ leads to $(3x+1)(x-1)=0$, or $x=-\frac13,\>1$. Check the extrema $f(-\frac13,\frac19) = \frac{32}{27}$ and $f(1,1) = 0$.

Therefore, the absolute minimum is $f(-2,4) = -9$ and the absolute maximum is $f(2,4) = 3$.

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  • $\begingroup$ How did you get f(-2, 4) and f(2, 4) $\endgroup$ – OGK Jan 26 at 4:53
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    $\begingroup$ @OGK - Note that $y=x^2$ and $y=4$ intersect at the points $(-2,4)$ and $(2,4)$. They are the corners on the boundary. $\endgroup$ – Quanto Jan 26 at 4:55
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Why you don't simply use the Lagrange method?

Btw, it seems you meant on or below y=x^2 and on or above y=4. In your setting, the function does not have a global max and min.

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