Maxima and Minima of Functions of Several Variables

Where the Slope Vanishes

Imagine you're standing on a hilly landscape, represented by $z = f(x,y)$. If you're truly at the top of a hill, then every direction you step leads downward. That means the slope in every direction must be zero at that exact point. In particular, the slope in the $x$-direction and the slope in the $y$-direction must both be zero:

Why is this necessary? Suppose $f$ has a relative maximum at $(x_0, y_0)$. Fix $y = y_0$ and look at the single-variable function $g(x) = f(x, y_0)$. This function has a maximum at $x_0$, so $g'(x_0) = f_x(x_0, y_0) = 0$. The same argument with $x = x_0$ fixed gives $f_y(x_0, y_0) = 0$. The reasoning for a minimum is identical.

Points where both partial derivatives vanish are called critical points. Every relative max and every relative min must occur at a critical point. But - and this is the crucial question - does a critical point have to be a max or min?

Think about $z = x^3$ in one variable: $z' = 0$ at $x = 0$, but that's neither a max nor a min - it's an inflection point. In two variables, even stranger things happen. Consider $z = x^2 - y^2$ at the origin: $f_x = 0$ and $f_y = 0$ at $(0,0)$, but it curves up in the $x$-direction and down in the $y$-direction. That's a saddle point - a critical point that's neither max nor min.

So we need a way to classify critical points. We need a test.

The Second Derivative Test

In one variable, the test is simple: at a critical point $x_0$, if $f''(x_0) > 0$ the curve bends upward (minimum), and if $f''(x_0) \lt 0$ it bends downward (maximum). We need to generalize this to two variables, where the surface can bend differently in different directions.

Here's the key idea. At a critical point $(x_0, y_0)$, the first directional derivative $\nabla_\alpha f$ is zero in every direction $\alpha$. So we look at the second directional derivative - the curvature of the surface along a line through the critical point in direction $\alpha$:

where we use the abbreviations:

$A = f_{xx}(x_0, y_0), \qquad B = f_{xy}(x_0, y_0), \qquad C = f_{yy}(x_0, y_0)$

This expression $P(\alpha) = A\cos^2\alpha + 2B\sin\alpha\cos\alpha + C\sin^2\alpha$ tells us how the surface curves in direction $\alpha$. For a maximum, we need $P(\alpha) \lt 0$ for every direction. For a minimum, $P(\alpha) > 0$ for every direction.

So when is $P(\alpha)$ always positive (or always negative)? There's a lovely algebraic argument. For $\alpha \ne \pm\pi/2$, we can factor out $\cos^2\alpha$ and write $P(\alpha) = \cos^2\alpha \cdot Q(\tan\alpha)$ where $Q(u) = Cu^2 + 2Bu + A$ is a quadratic in $u = \tan\alpha$. This quadratic has no real roots precisely when its discriminant $(2B)^2 - 4CA = 4(B^2 - AC) \lt 0$, i.e., when $AC - B^2 > 0$. In that case, $Q(u)$ has the same sign for all $u$, and that sign is determined by $A$ (just plug in $u = 0$).

This gives us the classification:

The trap here: students sometimes check $A > 0$ and $C > 0$ separately and conclude "minimum." That's not enough! You could have $A > 0$, $C > 0$, and still get a saddle point if $B$ is large enough to make $AC - B^2 \lt 0$. The cross-derivative $B = f_{xy}$ captures how the surface twists, and a strong enough twist overpowers the individual curvatures.

Worked example. Take $z = x^2 + 3xy + y^2 - 5x - 5y$. Setting $f_x = 2x + 3y - 5 = 0$ and $f_y = 3x + 2y - 5 = 0$, we solve: subtracting gives $-x + y = 0$, so $y = x$. Substitute back: $5x - 5 = 0$, giving $x = 1$, $y = 1$. The critical point is $(1, 1)$.

Now for the second derivatives: $f_{xx} = 2$, $f_{xy} = 3$, $f_{yy} = 2$. So $A = 2$, $B = 3$, $C = 2$. Check: $AC - B^2 = (2)(2) - (3)^2 = 4 - 9 = -5 \lt 0$. This is a saddle point - despite both $A > 0$ and $C > 0$ (positive curvature in each coordinate direction individually), the cross-term overwhelms them both.

Saddle Points: Seeing the Sign Change

A saddle point is beautiful and subtle. The surface passes through the critical point like a mountain pass - it curves upward along the ridge but downward along the valley. The second directional derivative $P(\alpha)$ literally changes sign as you rotate your direction.

Here's a classic example: $z = x^2 - y^2$. At the origin, $A = 2$, $B = 0$, $C = -2$, so $AC - B^2 = -4 \lt 0$ - a saddle. The directional second derivative is $P(\alpha) = 2\cos^2\alpha - 2\sin^2\alpha = 2\cos(2\alpha)$. Rotate the direction and watch the curvature swing from positive (upward) to negative (downward).

Left: the surface near the critical point. Right: a polar plot of $P(\alpha)$ - the second directional derivative as a function of direction. For a minimum, $P(\alpha) > 0$ everywhere (the polar curve stays in the positive region). For a saddle, it crosses zero and dips negative.

Notice how the level curves reveal the geometry too. At a max or min, level curves form closed ovals around the critical point. At a saddle, they form hyperbola-like curves - think of the contour lines around a mountain pass.

Absolute Extrema on Closed Regions

So far we've classified relative maxima and minima - local hilltops and valleys. But often we need the absolute largest or smallest value of $f$ on some region. Think of it this way: if you're optimizing the design of a container subject to constraints, you need the global best, not just a local one.

The key theorem (generalizing the familiar one-variable result) says:

The algorithm for finding them mirrors the one-variable case:

1. Find all critical points of $f$ inside the region (where $f_x = 0$ and $f_y = 0$).
2. Optimize $f$ on the boundary. Parametrize the boundary curve and reduce to a one-variable problem, or use Lagrange multipliers (§2.20).
3. Compare all values - the largest is the absolute max, the smallest is the absolute min.

Worked example (from Kaplan). Find the max and min of $z = x^2 + 2y^2 - x$ on the disk $x^2 + y^2 \le 1$.

Interior critical point: $f_x = 2x - 1 = 0$ and $f_y = 4y = 0$ give the critical point $(\frac{1}{2}, 0)$. Here $z = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$. We can verify it's a minimum: $A = 2$, $B = 0$, $C = 4$, so $AC - B^2 = 8 > 0$ and $A > 0$.

On the boundary $x^2 + y^2 = 1$: substitute $y^2 = 1 - x^2$ to get $z = x^2 + 2(1 - x^2) - x = 2 - x - x^2$ for $-1 \le x \le 1$. Setting $dz/dx = -1 - 2x = 0$ gives $x = -\frac{1}{2}$, so $y = \pm\frac{\sqrt{3}}{2}$, and $z = 2 + \frac{1}{2} - \frac{1}{4} = \frac{9}{4}$. At the endpoints: $x = 1$ gives $z = 0$; $x = -1$ gives $z = 2$.

Comparing all values: absolute min $= -\frac{1}{4}$ at $(\frac{1}{2}, 0)$, absolute max $= \frac{9}{4}$ at $(-\frac{1}{2}, \pm\frac{\sqrt{3}}{2})$.

Common mistake: forgetting the boundary entirely. Students find the interior critical point, classify it as a minimum, and declare it the answer. But the absolute max here is on the boundary - you'll miss it if you only look inside.

The surface $z = x^2 + 2y^2 - x$ over the unit disk, with the interior minimum (blue) and boundary maximum (red) marked.

Worked Example: Bounded Domain

Let's put the full algorithm to work. Consider the inverted paraboloid $z = 4 - x^2 - y^2$ on the closed disk $x^2 + y^2 \le 4$.

Predict first. This is a downward-opening paraboloid with vertex at the origin. We expect the peak to be at $(0,0)$ (where $z = 4$), and as we move outward, $z$ decreases. On the boundary circle $x^2 + y^2 = 4$, we get $z = 4 - 4 = 0$. So we should find a maximum at the origin and a minimum of zero on the boundary. Let's verify with calculus.

Step 1: Interior critical points. Set both partial derivatives to zero:

$f_x = -2x = 0, \qquad f_y = -2y = 0$

The only critical point is $(0, 0)$, where $z = 4$.

Step 2: Classify the critical point. The second derivatives are $A = f_{xx} = -2$, $B = f_{xy} = 0$, $C = f_{yy} = -2$. So:

$AC - B^2 = (-2)(-2) - 0^2 = 4 > 0, \qquad A = -2 < 0$

This confirms a local maximum at $(0, 0)$.

Step 3: Check the boundary. On $x^2 + y^2 = 4$, we have $z = 4 - 4 = 0$ everywhere. The boundary contributes a constant value of 0.

Result: Absolute maximum $= 4$ at $(0, 0)$. Absolute minimum $= 0$ on the entire boundary circle.

Worked Example: Unbounded Domain

What changes when there's no boundary at all? Consider $z = \dfrac{1}{1 + x^2 + y^2}$ on all of $\mathbb{R}^2$.

Predict first. This function is always positive (the denominator is always at least 1). At the origin, $z = 1$. As we move away from the origin, the denominator grows, so $z$ shrinks toward 0. We expect a maximum at the origin and no minimum - the function approaches 0 but never reaches it.

Step 1: Find critical points. Using the quotient rule:

$f_x = \dfrac{-2x}{(1 + x^2 + y^2)^2} = 0, \qquad f_y = \dfrac{-2y}{(1 + x^2 + y^2)^2} = 0$

Since the denominator is never zero, we need $x = 0$ and $y = 0$. The only critical point is $(0, 0)$, where $z = 1$.

Step 2: Classify. We need the second derivatives at the origin. At $(0, 0)$:

$A = f_{xx}(0,0) = -2, \qquad B = f_{xy}(0,0) = 0, \qquad C = f_{yy}(0,0) = -2$

$AC - B^2 = (-2)(-2) - 0 = 4 > 0, \qquad A = -2 < 0$

This is a local maximum.

Step 3: Behavior at infinity. As $(x, y) \to \infty$, $z \to 0^+$. But $z > 0$ everywhere - the value 0 is never actually achieved.

Result: Absolute maximum $= 1$ at $(0, 0)$. No absolute minimum - the infimum is 0, but it's never attained.

Worked Example: When Interior Fails, Try Lagrange

Sometimes the interior has no critical points at all, and the extrema live entirely on the boundary. Consider $z = x^2 + 2y$ on $x^2 + y^2 \le 4$.

Step 1: Look for interior critical points. Set the partial derivatives to zero:

$f_x = 2x = 0, \qquad f_y = 2 = 0$

The first equation gives $x = 0$, but the second equation $2 = 0$ is never satisfied. There are no interior critical points. The function $z = x^2 + 2y$ has no place in the interior where the slope vanishes in all directions - the $y$-direction always slopes upward. So the extrema must occur on the boundary.

Step 2: Optimize on the boundary using Lagrange multipliers. On the circle $g(x,y) = x^2 + y^2 - 4 = 0$, we set $\nabla f = \lambda \nabla g$:

$2x = \lambda \cdot 2x, \qquad 2 = \lambda \cdot 2y$

From the first equation: $2x(1 - \lambda) = 0$, so either $x = 0$ or $\lambda = 1$.

Case 1: $\lambda = 1$. The second equation gives $2 = 2y$, so $y = 1$. From the constraint: $x^2 + 1 = 4$, so $x^2 = 3$, giving $x = \pm\sqrt{3}$. At these points, $z = 3 + 2(1) = 5$.

Case 2: $x = 0$. From the constraint: $y^2 = 4$, so $y = \pm 2$. We get $z = 0 + 2(2) = 4$ at $(0, 2)$ and $z = 0 + 2(-2) = -4$ at $(0, -2)$.

Step 3: Compare all boundary values.

$z(\pm\sqrt{3}, 1) = 5, \qquad z(0, 2) = 4, \qquad z(0, -2) = -4$

Result: Absolute maximum $= 5$ at $(\pm\sqrt{3}, 1)$. Absolute minimum $= -4$ at $(0, -2)$.

Contour plot of $z = x^2 + 2y$ with the constraint circle $x^2 + y^2 = 4$. The red dots mark the maxima at $(\pm\sqrt{3}, 1)$ and the blue dot marks the minimum at $(0, -2)$.