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.


Along the boundary $y=4$,


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

Along the boundary $y=x^2$,


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 $f(2,4) = 3$.

| cite | improve this answer | |
  • $\begingroup$ How did you get f(-2, 4) and f(2, 4) $\endgroup$ – OGK Jan 26 at 4:53
  • 1
    $\begingroup$ @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

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.