Critical Points

Learning Objectives

Use partial derivatives to locate critical points for a function of two variables.

For functions of a single variable, we defined critical points as the values of the function when the derivative equals zero or does not exist. For functions of two or more variables, the concept is essentially the same, except for the fact that we are now working with partial derivatives.

Definition

Let $z=f(x, y)$ be a function of two variables that is defined on an open set containing the point $(x_0, y_0)$ . The point $(x_0, y_0)$ is called a critical point of a function of two variables $f$ if one of the two following conditions holds:

$f_x=(x_0, y_0)=f_y(x_0, y_0)=0$
Either $f_x(x_0, y_0)$ or $f_y(x_0, y_0)$ does not exist.

Example: Finding critical points

Find the critical points of each of the following functions:

a. $f(x,y)=\sqrt{4y^2-9x^2+24y+36x+36}$

b. $g(x,y)=x^2+2xy-4y^2+4x-6y+4$

Show Solution

a. First, we calculate $f_x(x, y)$ and $f_y(x, y)$ :

$\hspace{3cm}\begin{align} f_x(x,y)&=\frac12(-18x+36)(4y^2-9x^2+24y+36x+36)^{-1/2} \\ &=\frac{-9x+18}{\sqrt{4y^2-9x^2+24y+36x+36}} \\ f_y(x,y)&=\frac12(8y+24)(4y^2-9x^2+24y+36x+36)^{-1/2} \\ &=\frac{4y+12}{\sqrt{4y^2-9x^2+24y+36x+36}}. \end{align}$

Next, we set each of these expressions equal to zero:

$\hspace{6cm}\begin{align} \frac{-9x+18}{\sqrt{4y^2-9x^2+24y+36x+36}}&=0 \\ \frac{4y+12}{\sqrt{4y^2-9x^2+24y+36x+36}}&=0. \end{align}$

Then, multiply each equation by its common denominator:

$\hspace{10cm}\large{\begin{align} -9x+18&=0 \\ 4y+2&=0. \end{align}}$

Therefore, $x=2$ and $y=-3$ , so $(2, -3)$ is a critical point of $f$ .

We must also check for the possibility that the denominator of each partial derivative can equal zero, thus causing the partial derivative not to exist. Since the denominator is the same in each partial derivative, we need only do this once:

$\large{4y^{2}-9x^{2}+24y+36x+36=0}$

This equation represents a hyperbola. We should also note that the domain of $f$ consists of points satisfying the inequality

$\large{4y^{2}-9x^{2}+24y+36x+36\geq 0}$

Therefore, any points on the hyperbola are not only critical points, they are also on the boundary of the domain. To put the hyperbola in standard form, we use the method of completing the square:

$\hspace{6cm}\begin{align} 4y^{2}-9x^{2}+24y+36x+36&=0 \\ 4y^{2}-9x^{2}+24y+36x&=-36 \\ 4y^{2}+24y-9x^{2}+36x&=-36 \\ 4(y^2+6y)-9(x^2-4x)&=-36 \\ 4(y^2+6y+9)-9(x^2-4x+4)&=-36+36-36 \\ 4(y+3)^2-9(x-2)^2&=-36. \end{align}$

Dividing both sides by $-36$ puts the equation in standard form:

$\hspace{7.65cm}\begin{align} \frac{4(y+3)^2}{-36}-\frac{9(x-2)^2}{-36}&=1 \\ \frac{(y-2)^2}{4}-\frac{(x+3)^2}{9}&=1. \end{align}$

Notice that point $(2, -3)$ is the center of the hyperbola.

b. First, we calculate $g_x(x, y)$ and $g_y(x, y)$ :

$\hspace{7.5cm}\large{\begin{align} g_x(x,y)&=2x+2y+4 \\ g_y(x,y)&=2x-8y-6. \end{align}}$

Next, we set each of these expressions equal to zero, which gives a system of equations in $x$ and $y$ :

$\hspace{8cm}\large{\begin{align} 2x+2y+4&=0 \\ 2x-8y-6&=0. \end{align}}$

Subtracting the second equation from the first gives $10y+10=0$ , so $y=-1$ . Substituting this into the first equation gives $2x+2(-1)+4=0$ , so $x=-1$ . Therefore $(-1,-1)$ is a critical point of $g$ (Figure 1). There are no points in $\mathbb{R}^2$ that make either partial derivative not exist.

The function g(x, y) = x2 + 2xy – 4y2 + 4x – 6y + 4 is shown with critical point (–1, –1, 5). The critical point is located where the derivative in the x and y directions are both zero.

Figure 1. The function $\small{g(x,y)}$ has a critical point at $\small{(-1,-1,5)}$ .

try it

Find the critical point of the function $f(x, y)=x^{3}+2xy-2x-4y$ .

Show Solution

$(2, -5)$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 4.34” here (opens in new window).

The main purpose for determining critical points is to locate relative maxima and minima, as in single-variable calculus. When working with a function of one variable, the definition of a local extremum involves finding an interval around the critical point such that the function value is either greater than or less than all the other function values in that interval. When working with a function of two or more variables, we work with an open disk around the point.

Definition

Let $z=f(x, y)$ be a function of two variables that is defined and continuous on an open set containing the point $(x_0, y_0)$ . Then $f$ has a local maximum at $(x_0, y_0)$ if

$f(x_0, y_0)\geq f(x, y)$

for all points $(x, y)$ within some disk centered at $(x_0, y_0)$ . The number $f(x_0, y_0)$ is called a local maximum value. If the preceding inequality holds for every point $(x, y)$ in the domain of $f$ , then $f$ has a global maximum (also called an absolute maximum) at $(x_0, y_0)$ .

The function $f$ has a local minimum at $(x_0, y_0)$ if

$f(x_0, y_0)\leq f(x, y)$

for all points $(x, y)$ within some disk centered at $(x_0, y_0)$ . The number $f(x_0, y_0)$ is called a local minimum value. If the preceding inequality holds for every point $(x, y)$ in the domain of $f$ , then $f$ has a global minimum (also called an absolute minimum) at $(x_0, y_0)$ .

If $f(x_0, y_0)$ is either a local maximum or local minimum value, then it is called a local extremum (see the following figure).

The function z = the square root of (16 – x2 – y2) is shown, which is the upper hemisphere of radius 4 with center at the origin. In the xy plane, the circle with radius 4 and center at the origin is highlighted; it has equation x2 + y2 = 16.

Figure 2. The graph of $\small{z=\sqrt{16-x^{2}-y^{2}}}$ has a maximum value when $\small{(x,y)=(0,0)}$ It attains its minimum value at the boundary of its domain, which is the circle $\small{x^{2}+y^{2}=16}$ .

In Maxima and Minima, we showed that extrema of functions of one variable occur at critical points. The same is true for functions of more than one variable, as stated in the following theorem.

Theorem: Fermat’s theorem for functions of two variables

Let $z=f(x, y)$ be a function of two variables that is defined and continuous on an open set containing the point $(x_0, y_0)$ . Suppose $f_x$ and $f_y$ each exists at $(x_0, y_0)$ . If $f$ has a local extremum at $(x_0, y_0)$ , then $(x_0, y_0)$ is a critical point of $f$ .

Module 4: Differentiation of Functions of Several Variables

Learning Objectives

Definition

Example: Finding critical points

try it

Definition

Theorem: Fermat’s theorem for functions of two variables

Candela Citations