# Absolute Minimum and Maximum (region above, region below)

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 $$y=4$$,

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

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 $$f(2,4) = 3$$.

• How did you get f(-2, 4) and f(2, 4) – OGK Jan 26 at 4:53
• @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. – 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.