A critical point finds local highs and lows inside a region, but the highest point of a whole region is often on its edge. Think of a sloping field: its highest point is at a fence, where the gradient is not zero at all. As in Calculus 1, where you checked endpoints, the boundary must be checked too.
Key idea
A continuous function on a closed, bounded region has an absolute maximum and an absolute minimum. To find them:
- Find the critical points of $f$ inside the region and the values of $f$ there.
- On each piece of the boundary, write $f$ as a function of one variable and find its critical points.
- Include the corners (the endpoints of the boundary pieces).
- The largest value in the whole list is the absolute maximum; the smallest is the absolute minimum.
Why it works
Where can the maximum be? If it is at an interior point, $f$ is locally largest there, so $\nabla f = \mathbf{0}$: step 1 catches it. If it is on the boundary, it is in particular the largest value of $f$ along that edge, which is a one-variable problem: either a critical point of the edge function or an end of the edge, a corner. Steps 2 and 3 catch those. Nothing else is possible, so comparing the finite list is enough.
Worked examples
Worked example: On a square
Find the absolute extremes of $f(x, y) = x^2 + y^2 - 2x - 4y$ on ${}0 \le x \le 3$, ${}0 \le y \le 3$.
- Inside: $f_x = 2x - 2 = 0$, $f_y = 2y - 4 = 0$ at $(1, 2)$, where $f = -5$.
- Edge $y = 0$: $x^2 - 2x$, critical at $x = 1$: $f(1, 0) = -1$. Edge $x = 3$: $y^2 - 4y + 3$, critical at $y = 2$: $f(3, 2) = -1$.
- Edge $y = 3$: $x^2 - 2x - 3$, critical at $x = 1$: $f(1, 3) = -4$. Edge $x = 0$: $y^2 - 4y$, critical at $y = 2$: $f(0, 2) = -4$.
- Corners: $f(0, 0) = 0$, $f(3, 0) = 3$, $f(3, 3) = 0$, $f(0, 3) = -3$.
- Compare everything: largest ${}3$, smallest $-5$.
Answer: maximum ${}3$ at $(3, 0)$; minimum $-5$ at $(1, 2)$
Worked example: On a triangle
Find the absolute extremes of $f(x, y) = 2x^2 + y^2 - 4x$ on the triangle with vertices $(0, 0)$, $(2, 0)$, $(0, 2)$.
- Inside: $f_x = 4x - 4 = 0$ and $f_y = 2y = 0$ give $(1, 0)$, which is on the edge, not inside.
- Edge $y = 0$: ${}2x^2 - 4x$, critical at $x = 1$: $f = -2$. Edge $x = 0$: $y^2$, no interior critical point.
- Hypotenuse $y = 2 - x$: ${}2x^2 + (2 - x)^2 - 4x = 3x^2 - 8x + 4$, critical at $x = \frac{4}{3}$: $f = -\frac{4}{3}$.
- Corners: $f(0, 0) = 0$, $f(2, 0) = 0$, $f(0, 2) = 4$.
- Largest ${}4$, smallest $-2$.
Answer: maximum ${}4$ at $(0, 2)$; minimum $-2$ at $(1, 0)$
Common mistake
Skipping the boundary, or its corners. Setting $\nabla f = \mathbf{0}$ only finds interior candidates. In the first example the maximum, ${}3$, is at a corner, where neither partial derivative is zero. Each edge needs its own one-variable search, and every corner must be on the list, because the edge functions are only defined on closed intervals.