Local extrema and the second derivative test

In single-variable calculus you found the points where $f'(x) = 0$ and looked at the sign of $f''$. In two variables the idea is the same but with two differences: to find critical points you solve two equations at once, and the second derivative test is done not with a single number but with the discriminant $D$, which combines the second derivatives. A new type of point also appears: the saddle point, which behaves like a minimum in one direction and a maximum in another. In the exam this topic almost always comes as a “find and classify all critical points” question; the real points come from solving the system of equations completely.

Golden rules:

1. Critical point: $f_x = 0$ and $f_y = 0$. When solving the system, factor; do not lose roots by dividing.
2. Compute $D = f_{xx}f_{yy}-f_{xy}^2$ separately at each critical point.
3. If $D<0$ it is a saddle; if $D>0$ look at the sign of $f_{xx}$. If $D = 0$ the test is silent.
4. In the answer write the point, its type and the function value.

Definitions

If $f$ satisfies $f(x,y)\le f(a,b)$ at all points around $(a,b)$, then $(a,b)$ is a local maximum; if it satisfies $f(x,y)\ge f(a,b)$, it is a local minimum.

Theorem (first derivative). If $f$ has a local extremum at $(a,b)$ and the partial derivatives exist, then $f_x(a,b) = 0$ and $f_y(a,b) = 0$. That is, $\nabla f = \mathbf 0$ and the tangent plane is horizontal.

Points where $\nabla f = \mathbf 0$ (or where one of the partial derivatives is undefined) are called critical points. Every extremum is a critical point, but not every critical point is an extremum.

Saddle point. A critical point that is not an extremum. The classic example is $z = y^2-x^2$ (hyperbolic paraboloid): at the origin a maximum along the $x$ axis and a minimum along the $y$ axis. Like the middle of a horse saddle.

The second derivative test

Let $(a,b)$ be a critical point and the second derivatives be continuous.

$D = f_{xx}(a,b)\,f_{yy}(a,b)-\big[f_{xy}(a,b)\big]^2 = \begin{vmatrix}f_{xx}&f_{xy}\\f_{xy}&f_{yy}\end{vmatrix}$

Intuition. $D>0$ means the surface bends to the same side in all directions at the critical point; either a bowl (min) or an upside-down bowl (max). The curvature in any direction, for example $f_{xx}$, tells which. $D<0$ means that it bends up in some directions and down in others: a saddle. The $f_{xy}$ term accounts for curvature in mixed directions; even if $f_{xx}$ and $f_{yy}$ are both positive, if $f_{xy}$ is large the point can be a saddle.

PITFALL · saying min/max just because $f_{xx}$ and $f_{yy}$ have the same sign. $f = x^2+4xy+y^2$: $f_{xx} = 2>0$, $f_{yy} = 2>0$, but $f_{xy} = 4$ and $D = 4-16 = -12<0$. The origin is a saddle. Along the line $y = -x$, $f = -2x^2$, going down. Do not decide without computing $D$.

Solved examples

Example 1 (a single critical point). $f(x,y) = x^2+2y^2-4x+4y+1$.

f_x = 2x - 4 = 0    →  x = 2
f_y = 4y + 4 = 0    →  y = -1
f_xx = 2, f_yy = 4, f_xy = 0,  D = 8 > 0, f_xx > 0  →  local min
f(2,-1) = 4 + 2 - 8 - 4 + 1 = -5

Example 2 (two critical points). $f(x,y) = xy-\dfrac{x^3}{3}-\dfrac{y^2}{2}$.

$f_x = y-x^2 = 0$ and $f_y = x-y = 0$. From the second, $y = x$; plug into the first: $x-x^2 = x(1-x) = 0$. The points are $(0,0)$ and $(1,1)$.

Second derivatives: $f_{xx} = -2x$, $f_{yy} = -1$, $f_{xy} = 1$, $D = 2x-1$.

TIP · do not lose roots. If you divide the equation $x-x^2 = 0$ by $x$ and write $x = 1$, you lose $(0,0)$. Always factor: $x(1-x) = 0$. The equations $x^4 = x$, $x^9 = x$ in Q4 and Q6 are the same trap.

Strategies for solving systems

Example 3 ($D = 0$). $f = x^4-y^4$, at the origin all second derivatives are $0$, $D = 0$. The test is silent. Look at paths: on the $x$ axis $f = x^4\ge0$, on the $y$ axis $f = -y^4\le0$. In every neighborhood of the origin there are both positive and negative values: a saddle. In the same situation $x^4+y^4$ is a minimum and $-(x^4+y^4)$ a maximum. So when $D = 0$ anything can happen.

Application problems

Optimization problems usually have three variables and one constraint. Use the constraint to eliminate a variable, reduce to a two-variable function, then apply the test on this page.

1) What are you optimizing? (area, volume, distance, cost) → formula
2) Constraint: xyz = V, x + y + z = k, the point is on a plane ...  → pull out z
3) Build the two-variable function, find the critical point
4) Confirm with D that it is a min/max (or physically it is the only candidate)
5) Compute the third variable and the requested value

The distance trick. When looking for the closest point to a point, minimize the square of the distance, not $\sqrt{\ldots}$. Since the root is an increasing function, the minimum is at the same point and the derivatives come out much cleaner.

Next week we will see how to solve the same problems without eliminating variables, with Lagrange multipliers, and how to find absolute extrema on closed regions.

Common mistakes