Skip to main content
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 2.5: Maxima and Minima

• • Contributed by Michael Corral
• Professor (Mathematics) at Schoolcraft College

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

The gradient can be used to find extreme points of real-valued functions of several variables, that is, points where the function has a local maximum or local minimum. We will consider only functions of two variables; functions of three or more variables require methods using linear algebra.

Definition 2.7

Let $$f (x, y)$$ be a real-valued function, and let $$(a,b)$$ be a point in the domain of $$f$$ . We say that $$f$$ has a local maximum at $$(a,b)$$ if $$f (x, y) \le f (a,b)$$ for all $$(x, y)$$ inside some disk of positive radius centered at $$(a,b)$$, i.e. there is some sufficiently small $$r > 0$$ such that $$f (x, y) \le f (a,b)\text{ for all }(x, y)$$ for which $$(x− a)^2 +(y− b)^2 < r^2$$ .

Likewise, we say that $$f$$ has a local minimum at $$(a,b)\text{ if }f (x, y) \gt f (a,b)\text{ for all }(x, y)$$ inside some disk of positive radius centered at $$(a,b)$$.

If $$f (x, y) \le f (a,b)\text{ for all }(x, y)$$ in the domain of $$f$$ , then $$f$$ has a global maximum at $$(a,b)$$. If $$f (x, y) \ge f (a,b)\text{ for all }(x, y)\text{ in the domain of }f \text{, then }f$$ has a global minimum at $$(a,b)$$.

Suppose that $$(a,b)$$ is a local maximum point for $$f (x, y)$$, and that the first-order partial derivatives of $$f$$ exist at $$(a,b)$$. We know that $$f (a,b)$$ is the largest value of $$f (x, y)\text{ as }(x, y)$$ goes in all directions from the point $$(a,b)$$, in some sufficiently small disk centered at $$(a,b)$$. In particular, $$f (a,b)$$ is the largest value of $$f$$ in the $$x$$ direction (around the point $$(a,b)$$), that is, the single-variable function $$g(x) = f (x,b)$$ has a local maximum at $$x = a$$. So we know that $$g ′ (a) = 0$$. Since $$g ′ (x) = \dfrac{∂f}{∂x} (x,b)\text{, then }\dfrac{∂f}{∂x} (a,b) = 0$$. Similarly, $$f (a,b)$$ is the largest value of $$f$$ near $$(a,b)$$ in the $$y$$ direction and so $$\dfrac{∂f}{∂y} (a,b) = 0$$. We thus have the following theorem:

Theorem 2.5

Let $$f (x, y)$$ be a real-valued function such that both $$\dfrac{∂f}{∂x} (a,b)$$ and $$\dfrac{∂f}{∂y} (a,b)$$ exist. Then a necessary condition for $$f (x, y)$$ to have a local maximum or minimum at $$(a,b)$$ is that $$\nabla f (a,b) = \textbf{0}$$.

Note: Theorem 2.5 can be extended to apply to functions of three or more variables.

A point $$(a,b)$$ where $$\nabla f (a,b) = \textbf{0}$$ is called a critical point for the function $$f (x, y)$$. So given a function $$f (x, y)$$, to find the critical points of $$f$$ you have to solve the equations $$\dfrac{∂f}{∂x} (x, y) = 0\text{ and }\dfrac{∂f}{∂y} (x, y) = 0$$ simultaneously for $$(x, y)$$. Similar to the single-variable case, the necessary condition that $$\nabla f (a,b) = \textbf{0}$$ is not always sufficient to guarantee that a critical point is a local maximum or minimum.

Example 2.18

The function $$f (x, y) = x y$$ has a critical point at $$(0,0)$$: $$\dfrac{∂f}{∂x} = y = 0 \Rightarrow y = 0$$, and $$\dfrac{∂f}{∂y} = x = 0 \Rightarrow x = 0$$, so $$(0,0)$$ is the only critical point. But clearly $$f$$ does not have a local maximum or minimum at $$(0,0)$$ since any disk around $$(0,0)$$ contains points $$(x, y)$$ where the values of $$x$$ and $$y$$ have the same sign (so that $$f (x, y) = x y \gt 0 = f (0,0))$$ and different signs (so that $$f (x, y) = x y < 0 = f (0,0))$$. In fact, along the path $$y = x$$ in $$\mathbb{R}^2$$ , $$f (x, y) = x^2$$ , which has a local minimum at $$(0,0)$$, while along the path $$y = −x$$ we have $$f (x, y) = −x^2$$ , which has a local maximum at $$(0,0)$$. So $$(0,0)$$ is an example of a saddle point, i.e. it is a local maximum in one direction and a local minimum in another direction. The graph of $$f (x, y)$$ is shown in Figure 2.5.1, which is a hyperbolic paraboloid. Figure 2.5.1 $$f (x, y) = x y$$, saddle point at (0,0)

The following theorem gives sufficient conditions for a critical point to be a local maximum or minimum of a smooth function (i.e. a function whose partial derivatives of all orders exist and are continuous), which we will not prove here.

Theorem 2.6

Let $$f (x, y)$$ be a smooth real-valued function, with a critical point at $$(a,b)$$ (i.e. $$\nabla f (a,b) = 0$$). Define

$\nonumber D = \dfrac{∂^2 f}{∂x^2} (a,b) \dfrac{∂^2 f}{∂y^2}(a,b)− \left (\dfrac{∂^2 f}{∂y∂x} (a,b)\right )^2$

Then

1. if $$D > 0 \text{ and }\dfrac{∂^2 f}{∂x^2} (a,b) > 0$$, then $$f$$ has a local minimum at $$(a,b)$$
2. if $$D > 0 \text{ and }\dfrac{∂^2 f}{∂x^2} (a,b) < 0$$, then $$f$$ has a local maximum at $$(a,b)$$
3. if $$D < 0 \text{, then }f$$ has neither a local minimum nor a local maximum at $$(a,b)$$
4. if $$D = 0$$, then the test fails

If condition (c) holds, then $$(a,b)$$ is a saddle point. Note that the assumption that $$f (x, y)$$ is smooth means that

$D = \begin{vmatrix} \dfrac{∂^2 f}{∂x^2} (a,b) & \dfrac{∂^2 f}{∂y∂x} (a,b) \\ \dfrac{∂^2 f}{∂x∂y} (a,b) & \dfrac{∂^2 f}{∂y^2} (a,b) \\ \end{vmatrix}$

since $$\dfrac{∂^2 f}{∂y∂x} = \dfrac{∂^2 f}{∂x∂y}$$ . Also, if $$D > 0$$ then $$\dfrac{∂^2 f}{∂x^2} (a,b) \dfrac{∂^2 f}{∂y^2} (a,b) = D + \left ( \dfrac{∂^2 f}{∂y∂x} (a,b)\right )^2 > 0$$, and so $$\dfrac{∂^2 f}{∂x^2} (a,b)\text{ and }\dfrac{∂^2 f}{∂y^2} (a,b)$$ have the same sign. This means that in parts (a) and (b) of the theorem one can replace $$\dfrac{∂^2 f}{∂x^2} (a,b)$$ by $$\dfrac{∂^2 f}{∂y^2} (a,b)$$ if desired.

Example 2.19

Find all local maxima and minima of $$f (x, y) = x^2 + x y+ y^2 −3x$$.

Solution

First find the critical points, i.e. where $$\nabla f = \textbf{0}$$. Since

$\nonumber \dfrac{∂f}{∂x} = 2x+ y−3 \text{ and }\dfrac{∂f}{∂y} = x+2y$

then the critical points $$(x, y)$$ are the common solutions of the equations

\nonumber \begin{align} 2x+ y−3 &= 0 \\ \nonumber x +2y &= 0 \end{align}

which has the unique solution $$(x, y) = (2,−1)$$. So (2,−1) is the only critical point.

To use Theorem 2.6, we need the second-order partial derivatives:

$\nonumber \dfrac{∂^2 f}{∂x^2} = 2 ,\quad \dfrac{∂^2 f}{∂y^2} = 2 ,\quad \dfrac{∂^2 f}{∂y∂x} = 1$

and so

$\nonumber D = \dfrac{∂^2 f}{∂x^2} (2,−1) \dfrac{∂^2 f}{∂y^2} (2,−1)− \left ( \dfrac{∂^2 f}{∂y∂x} (2,−1)\right )^2 = (2)(2)−1^2 = 3 > 0$

and $$\dfrac{∂^2 f}{∂x^2} (2,−1) = 2 > 0$$. Thus, (2,−1) is a local minimum.

Example 2.20

Find all local maxima and minima of $$f (x, y) = x y− x^3 − y^2$$ .

Solution

First find the critical points, i.e. where $$\nabla f = \textbf{0}$$. Since

$\nonumber \dfrac{∂f}{∂x} = y−3x^2 \text{ and }\dfrac{∂f}{∂y} = x−2y$

then the critical points $$(x, y)$$ are the common solutions of the equations

\nonumber \begin{align} y−3x^2 &= 0 \\ \nonumber x−2y &= 0 \end{align}

The first equation yields $$y = 3x^2$$, substituting that into the second equation yields $$x−6x^2 = 0$$, which has the solutions $$x = 0 \text{ and }x = \dfrac{1}{6}$$ . So $$x = 0 \Rightarrow y = 3(0) = 0 \text{ and }x = \dfrac{1}{6} \Rightarrow y = 3 \left (\dfrac{1}{6} \right )^2 = \dfrac{1}{12}$$ . So the critical points are $$(x, y) = (0,0)$$ and $$(x, y) = \left ( \dfrac{1}{6} , \dfrac{1}{12} \right )$$.

To use Theorem 2.6, we need the second-order partial derivatives:

$\nonumber \dfrac{∂^2 f}{∂x^2} = −6x ,\quad \dfrac{∂^2 f}{∂y^2} = −2 ,\quad \dfrac{∂^2 f}{∂y∂x} = 1$

So

$\nonumber D = \dfrac{∂^2 f}{∂x^2} (0,0) \dfrac{∂^2 f}{∂y^2} (0,0)− \left (\dfrac{∂^2 f}{∂y∂x} (0,0)\right )^2 = (−6(0))(−2)−1^2 = −1 < 0$

and thus $$(0,0)$$ is a saddle point. Also,

$\nonumber D = \dfrac{∂^2 f}{∂x^2} \left ( \dfrac{1}{6} ,\dfrac{1}{12 } \right ) \dfrac{∂^2 f}{∂y^2} \left (\dfrac{1}{6} , \dfrac{1}{12}\right ) − \left (\dfrac{∂^2 f}{∂y∂x} \left ( \dfrac{1}{6} , \dfrac{1}{12}\right ) \right)^2 = (−6 \left (\dfrac{1}{6}\right )) (−2)−1^2 = 1 > 0$

and $$\dfrac{∂^2 f}{∂x^2}\left (\dfrac{1}{6} , \dfrac{1}{12} \right ) = −1 < 0$$. Thus, $$\left (\dfrac{1}{6} , \dfrac{1}{12}\right )$$ is a local maximum.

Example 2.21

Find all local maxima and minima of $$f (x, y) = (x−2)^4 +(x−2y)^2$$ .

Solution

First find the critical points, i.e. where $$\nabla f = \textbf{0}$$. Since

$\nonumber \dfrac{∂f}{∂x} = 4(x−2)^3 +2(x−2y) \text{ and }\dfrac{∂f}{∂y} = −4(x−2y)$

then the critical points $$(x, y)$$ are the common solutions of the equations

\nonumber \begin{align} 4(x−2)^3 +2(x−2y) &= 0 \\ \nonumber −4(x−2y) &= 0 \end{align}

The second equation yields $$x = 2y$$, substituting that into the first equation yields $$4(2y−2)^3 = 0$$, which has the solution $$y = 1$$, and so $$x = 2(1) = 2$$. Thus, $$(2,1)$$ is the only critical point.

To use Theorem 2.6, we need the second-order partial derivatives:

$\nonumber \dfrac{∂^2 f}{∂x^2} = 12(x−2)^2 +2,\quad \dfrac{∂^2 f}{∂y^2} = 8 ,\quad \dfrac{∂^2 f}{∂y∂x} = −4$

So

$\nonumber D = \dfrac{∂^2 f}{∂x^2} (2,1) \dfrac{∂^2 f}{∂y^2} (2,1)− \left (\dfrac{∂^2 f}{∂y∂x} (2,1)\right )^2 = (2)(8)−(−4)^2 = 0$

and so the test fails. What can be done in this situation? Sometimes it is possible to examine the function to see directly the nature of a critical point. In our case, we see that $$f (x, y) \ge 0$$ for all $$(x, y)$$, since $$f (x, y)$$ is the sum of fourth and second powers of numbers and hence must be nonnegative. But we also see that $$f (2,1) = 0$$. Thus $$f (x, y) \ge 0 = f (2,1)$$ for all $$(x, y)$$, and hence $$(2,1)$$ is in fact a global minimum for $$f$$ .

Example 2.22

Find all local maxima and minima of $$f (x, y) = (x^2 + y^2 )e^{−(x^2+y^2)}$$ .

Solution

First find the critical points, i.e. where $$\nabla f = 0$$. Since

\nonumber \begin{align} \dfrac{∂f}{∂x} &= 2x(1−(x^2 + y^2 ))e^{−(x^2+y^2)} \\ \nonumber \dfrac{∂f}{∂y} &= 2y(1−(x^2 + y^2))e^{−(x^2+y^2)} \end{align}

then the critical points are $$(0,0)$$ and all points $$(x, y)$$ on the unit circle $$x^2 + y^2 = 1$$.

To use Theorem 2.6, we need the second-order partial derivatives:

\nonumber \begin{align} \dfrac{∂^2 f}{∂x^2} &= 2[1−(x^2 + y^2 )−2x^2 −2x^2 (1−(x^2 + y^2))]e^{−(x^2+y^2)} \\ \nonumber \dfrac{∂^2 f}{∂y^2} &= 2[1−(x^2 + y^2 )−2y^2 −2y^2 (1−(x^2 + y^2))]e^{−(x^2+y^2)} \\ \nonumber \dfrac{∂^2 f}{∂y∂x} &= −4x y[2−(x^2 + y^2 )]e^{−(x^2+y^2)} \end{align}

At $$(0,0)$$, we have $$D = 4 > 0$$ and $$\dfrac{∂^2 f}{∂x^2} (0,0) = 2 > 0$$, so $$(0,0)$$ is a local minimum. However, for points $$(x, y)$$ on the unit circle $$x^2 + y^2 = 1$$, we have

$\nonumber D = (−4x^2 e^{−1} )(−4y^2 e^{−1} )−(−4x ye^{−1} )^2 = 0$

and so the test fails. If we look at the graph of $$f (x, y)$$, as shown in Figure 2.5.2, it looks like we might have a local maximum for $$(x, y)$$ on the unit circle $$x^2 + y^2 = 1$$. If we switch to using polar coordinates $$(r,\theta )$$ instead of $$(x, y)$$ in $$\mathbb{R}^2$$ , where $$r^2 = x^2+y^2$$, then we see that we can write $$f (x, y)$$ as a function $$g(r)$$ of the variable $$r$$ alone: $$g(r) = r^2 e^{−r^2}$$ . Then $$g ′ (r) = 2r(1 − r^2 )e^{−r^2}$$ , so it has a critical point at $$r = 1$$, and we can check that $$g′′(1) = −4e^{−1} < 0$$, so the Second Derivative Test from single-variable calculus says that $$r = 1$$ is a local maximum. But $$r = 1$$ corresponds to the unit circle $$x^2 + y^2 = 1$$. Thus, the points $$(x, y)$$ on the unit circle $$x^2 + y^2 = 1$$ are local maximum points for $$f$$ . Figure 2.5.2 $$f (x, y) = (x^ 2 + y^ 2 )e^{ −(x^ 2+y^ 2 )}$$