Critical Points and Local Extrema: Learn It 1

Giselle Miller

Critical Points and Local Extrema: Learn It 1

Find critical points using partial derivatives
Apply the second derivative test
Find absolute extrema on closed regions

Critical Points

One of the most useful applications of derivatives for single-variable functions is finding maximum and minimum values. This application is equally important for functions of two or more variables. The main ideas—finding critical points and using derivative tests—are still valid, but more independent variables lead to more possible outcomes.

For functions of a single variable, critical points occur where the derivative equals zero or does not exist. For functions of two or more variables, the concept is the same, except we work with partial derivatives.

critical point

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

[latex]f_x=(x_0, y_0)=f_y(x_0, y_0)=0[/latex]
Either [latex]f_x(x_0, y_0)[/latex] or [latex]f_y(x_0, y_0)[/latex] does not exist.

Find the critical points of each of the following functions:

[latex]f(x,y)=\sqrt{4y^2-9x^2+24y+36x+36}[/latex]
[latex]g(x,y)=x^2+2xy-4y^2+4x-6y+4[/latex]

First, we calculate [latex]f_x(x, y)[/latex] and [latex]f_y(x, y)[/latex]:[latex]\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}[/latex]
Next, we set each of these expressions equal to zero:

[latex]\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}[/latex]

Then, multiply each equation by its common denominator:

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

Therefore, [latex]x=2[/latex] and [latex]y=-3[/latex], so [latex](2, -3)[/latex] is a critical point of [latex]f[/latex].

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:

[latex]\large{4y^{2}-9x^{2}+24y+36x+36=0}[/latex]

This equation represents a hyperbola. We should also note that the domain of [latex]f[/latex] consists of points satisfying the inequality

[latex]\large{4y^{2}-9x^{2}+24y+36x+36\geq 0}[/latex]

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:

[latex]\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}[/latex]

Dividing both sides by [latex]-36[/latex] puts the equation in standard form:

[latex]\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}[/latex]

Notice that point [latex](2, -3)[/latex] is the center of the hyperbola.
First, we calculate [latex]g_x(x, y)[/latex] and [latex]g_y(x, y)[/latex]:[latex]\hspace{7.5cm}\large{\begin{align} g_x(x,y)&=2x+2y+4 \\ g_y(x,y)&=2x-8y-6. \end{align}}[/latex]
Next, we set each of these expressions equal to zero, which gives a system of equations in [latex]x[/latex] and [latex]y[/latex]:

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

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

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

Local Extrema

The main purpose for determining critical points is to locate relative maxima and minima, just 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 other function values in that interval. For functions of two or more variables, we work with an open disk around the point.

local extrema

Let [latex]z=f(x, y)[/latex] be a function of two variables that is defined and continuous on an open set containing the point [latex](x_0, y_0)[/latex].

[latex]\\[/latex]

Local Maximum: The function [latex]f[/latex] has a local maximum at [latex](x_0, y_0)[/latex] if
[latex]f(x_0, y_0)\geq f(x, y)[/latex]

for all points [latex](x, y)[/latex] within some disk centered at [latex](x_0, y_0)[/latex].
- The number [latex]f(x_0, y_0)[/latex] is called a local maximum value. If this inequality holds for every point [latex](x, y)[/latex] in the domain of [latex]f[/latex], then [latex]f[/latex] has a global maximum (also called an absolute maximum) at [latex](x_0, y_0)[/latex].
Local Minimum: The function [latex]f[/latex] has a local minimum at [latex](x_0, y_0)[/latex] if
[latex]f(x_0, y_0)\leq f(x, y)[/latex]

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

If [latex]f(x_0, y_0)[/latex] is either a local maximum or local minimum value, then it is called a local extremum.

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

Just as with functions of one variable, extrema of functions of more than one variable occur at critical points.

Fermat’s theorem for functions of two variables

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