Unconstrained Multivariable Extrema

Peaks, Pits, and Saddles

To find extrema of $f(x,y)$ without constraints:

Step 1 — First-order conditions: set $f_x = 0$ and $f_y = 0$. Solutions are critical points — candidates for maxima, minima, or saddle points.

Step 2 — Second-order conditions: the Hessian matrix classifies each critical point:

$$H = \begin{pmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{pmatrix}$$

Compute the determinant $D = f_{xx} f_{yy} - (f_{xy})^2$:

• $D > 0$ and $f_{xx} < 0$: local maximum (the surface curves downward in all directions).
• $D > 0$ and $f_{xx} > 0$: local minimum (curves upward in all directions).
• $D < 0$: saddle point (curves up in one direction, down in another — a mountain pass).
• $D = 0$: inconclusive — higher-order analysis needed.

$$D = f_{xx} f_{yy} - f_{xy}^2 \qquad \begin{cases} D>0, f_{xx}<0 \Rightarrow \text{max} \\ D>0, f_{xx}>0 \Rightarrow \text{min} \\ D<0 \Rightarrow \text{saddle} \end{cases}$$

Economic application: a firm chooses output level $q$ and advertising expenditure $a$ to maximise profit $\pi(q,a)$. The FOCs give the candidate; the Hessian confirms it is a maximum (profit function should be concave — diminishing returns to both quantity and advertising). If the Hessian says saddle, the firm is at a pass, not a peak — increasing one variable while decreasing the other would improve profit.

The Hessian verdicts

Common pitfall: Checking $f_{xx} < 0$ and $f_{yy} < 0$ separately and declaring a maximum. The cross-term can betray you: with a large $f_{xy}$, the surface can still be a saddle. Only the determinant test sees the interaction.