Extremums

Revise Extremums from S1.

Theorems

Local extremum

Suppose $f$ has a local maximum or minimum at $(a,b)$ and the first order partial derivatives of $f$ exist there. Then $f_x (a, b) = 0$ and $f_y (a, b) = 0$.

Proof Hint

Fix 1 variable and apply the first derivative test.

Extreme Value Theorem

If $f$ is continuous on a closed, bounded set $D \subseteq \mathbb{R}^2$ then $f$ has an absolute maximum and absolute minimum value in $D$.

Implicit Function Theorem

Suppose $f(x, y) ∈ C^1$, $f_y(a, b) \neq 0$ and $f(a, b) = c$. Then there exists a unique function $y = g(x) ∈ C^1$ defined on a neighbourhood of $(a, b)$ with:

$$g'(x) = − \frac{f_x(x,g(x))} {f_y(x,g(x))}\;\; \text{s.t.}\;\; f(x, g(x)) = c$$

Second Derivatives Test

Suppose the second partial derivatives of $f$ are continuous on a disk with center $(a, b)$, and suppose that $f_x (a, b) = 0$ and $f_y (a, b) = 0$. Let:

$$D = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix} = D(a, b) = f_{xx}(a, b)f_{yy} (a, b) − f_{xy}^2(a, b)$$

• If $D=0$, then no information.
• If $D > 0 \land f_{xx} (a, b) > 0$, then $f(a, b)$ is a local minimum.
• If $D > 0 \land f_{xx} (a, b) < 0$, then $f(a, b)$ is a local maximum.
• If $D < 0$ then $f(a, b)$ is not a local maximum or minimum.

Saddle point

A critical point in which it is not a local maximum or local minimum.

Closed interval method

A method for finding absolute minimum and absolute maximum values of $f$ which is continuous on a closed, bounded set $D \subseteq \mathbb{R}^2$.

1. Find the values of $f$ at the critical points of $f$ in the interior of $D$.
2. Find the extreme values of $f$ on the boundary of $D$.
3. Maximum and minimum values from step 1 and 2 are the absolute maximum and absolute minimum.