Absolute Maxima and Minima

Learning Objectives

Examine critical points and boundary points to find absolute maximum and minimum values for a function of two variables.

When finding global extrema of functions of one variable on a closed interval, we start by checking the critical values over that interval and then evaluate the function at the endpoints of the interval. We briefly revisit the single-variable version below.

Recall: Extreme Value Theorem (single-variable version)

A continuous function [latex]f(x)[/latex] on a closed interval [latex][a,b][/latex] attains an absolute maximum value at some point of [latex]D[/latex] and an absolute minimum value at some point of [latex]D[/latex]. These values are, respectively, the largest and smallest values found among the following:

The values of [latex]f[/latex] at the critical values of [latex]f[/latex] within the interval [latex][a,b][/latex].
The values of [latex]f[/latex] at the endpoints of [latex][a,b][/latex].

When working with a function of two variables, the closed interval is replaced by a closed, bounded set. A set is bounded if all the points in that set can be contained within a ball (or disk) of finite radius. First, we need to find the critical points inside the set and calculate the corresponding critical values. Then, it is necessary to find the maximum and minimum value of the function on the boundary of the set. When we have all these values, the largest function value corresponds to the global maximum and the smallest function value corresponds to the absolute minimum. First, however, we need to be assured that such values exist. The following theorem does this.

Theorem: Extreme valUE theorem

A continuous function [latex]f(x, y)[/latex] on a closed and bounded set [latex]D[/latex] in the plane attains an absolute maximum value at some point of [latex]D[/latex] and an absolute minimum value at some point of [latex]D[/latex].

Now that we know any continuous function [latex]f[/latex] defined on a closed, bounded set attains its extreme values, we need to know how to find them.

Theorem: Extreme valUE theorem

Assume [latex]z=f(x, y)[/latex] is a differentiable function of two variables defined on a closed, bounded set [latex]D[/latex]. Then [latex]f[/latex] will attain the absolute maximum value and the absolute minimum value, which are, respectively, the largest and smallest values found among the following:

The values of [latex]f[/latex] at the critical points of [latex]f[/latex] in [latex]D[/latex].
The values of [latex]f[/latex] on the boundary of [latex]D[/latex].

The proof of this theorem is a direct consequence of the extreme value theorem and Fermat’s theorem. In particular, if either extremum is not located on the boundary of [latex]D[/latex], then it is located at an interior point of [latex]D[/latex]. But an interior point [latex](x_0, y_0)[/latex] of [latex]D[/latex] that’s an absolute extremum is also a local extremum; hence, [latex](x_0, y_0)[/latex] is a critical point of [latex]f[/latex] by Fermat’s theorem. Therefore the only possible values for the global extrema of [latex]f[/latex] on [latex]D[/latex] are the extreme values of [latex]f[/latex] on the interior or boundary of [latex]D[/latex].

Problem-solving strategy: using the second derivative test for functions of two variables

Let [latex]z=f(x, y)[/latex] be a continuous function of two variables defined on a closed, bounded set [latex]D[/latex], and assume [latex]f[/latex] is differentiable on [latex]D[/latex]. To find the absolute maximum and minimum values of [latex]f[/latex] on [latex]D[/latex], do the following:

Determine the critical points of [latex]f[/latex] on [latex]D[/latex].
Calculate [latex]f[/latex] at each of these critical points.
Determine the maximum and minimum values of [latex]f[/latex] on the boundary of its domain.
The maximum and minimum values of [latex]f[/latex] will occur at one of the values obtained in steps 2 and 3.

Finding the maximum and minimum values of [latex]f[/latex] on the boundary of [latex]D[/latex] can be challenging. If the boundary is a rectangle or set of straight lines, then it is possible to parameterize the line segments and determine the maxima on each of these segments, as seen in Example: Finding Absolute Extrema. The same approach can be used for other shapes such as circles and ellipses.

If the boundary of the set [latex]D[/latex] is a more complicated curve defined by a function [latex]g(x, y)=c[/latex] for some constant [latex]c[/latex], and the first-order partial derivatives of [latex]g[/latex] exist, then the method of Lagrange multipliers can prove useful for determining the extrema of [latex]f[/latex] on the boundary. The method of Lagrange multipliers is introduced in Lagrange Multipliers.

Example: Finding Absolute extrema

Use the problem-solving strategy for finding absolute extrema of a function to determine the absolute extrema of each of the following functions:

[latex]f(x, y)=x^{2}-2xy+4y^{2}-4x-2y+24[/latex] on the domain defined by [latex]0\leq x\leq 4[/latex] and [latex]0\leq y\leq 2[/latex].
[latex]g(x, y)=x^{2}+y^{2}+4x-6y[/latex] on the domain defined by [latex]x^{2}+y^{2}\leq 16[/latex].

Show Solution

a. Using the problem-solving strategy, step 1 involves finding the critical points of [latex]f[/latex] on its domain. Therefore, we first calculate [latex]f_x(x, y)[/latex] and [latex]f_y(x, y)[/latex], then set them each equal to zero:

[latex]\hspace{8cm}\begin{align} f_x(x, y)&=2x-2y-4 \\ f_y(x, y)&=-2x+8y-2. \end{align}[/latex]

Setting them equal to zero yields the system of equations

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

The solution to this system is [latex]x=3[/latex] and [latex]y=1[/latex]. Therefore [latex](3, 1)[/latex] is a critical point of [latex]f[/latex]. Calculating [latex]f(3, 1)[/latex] gives [latex]f(3, 1)=17[/latex].
The next step involves finding the extrema of [latex]f[/latex] on the boundary of its domain. The boundary of its domain consists of four line segments as shown in the following graph:

A rectangle is drawn in the first quadrant with one corner at the origin, horizontal length 4, and height 2. This rectangle is marked D, and the sides are marked in counterclockwise order from the side overlapping the x axis L1, L2, L3, and L4.

Figure 1. Graph of the domain of the function [latex]\small{f(x,y)=x^{2}-2xy+4y^{2}-4x-2y+24}[/latex].

[latex]L_1[/latex] is the line segment connecting [latex](0, 0)[/latex] and [latex](4, 0)[/latex], and it can be parameterized by the equations [latex]x(t)=t[/latex], [latex]y(t)=0[/latex] for [latex]0\leq t\leq 4[/latex]. Define [latex]g(t)=f(x(t), y(t))[/latex]. This gives [latex]g(t)=t^{2}-4t+24[/latex]. Differentiating [latex]g[/latex] leads to [latex]g'(t)=2t-4[/latex]. Therefore, [latex]g[/latex] has a critical value at [latex]t=2[/latex], which corresponds to the point [latex](2, 0)[/latex]. Calculating [latex]f(2, 0)[/latex] gives the [latex]z[/latex]-value [latex]20[/latex].

[latex]L_2[/latex] is the line segment connecting [latex](4, 0)[/latex] and [latex](4, 2)[/latex], and it can be parameterized by the equations [latex]x(t)=4[/latex], [latex]y(t)=t[/latex] for [latex]0\leq t\leq 2[/latex]. Again, define [latex]g(t)=f(x(t), y(t))[/latex]. This gives [latex]g(t)=4t^{2}-10t+24[/latex]. Then, [latex]g'(t)=8t-10[/latex]. [latex]g[/latex] has a critical value at [latex]t=\frac{5}{4}[/latex], which corresponds to the point [latex](4, \frac{5}{4})[/latex]. Calculating [latex]f(4, \frac{5}{4})[/latex] gives the [latex]z[/latex]-value [latex]17.75[/latex].

[latex]L_3[/latex] is the line segment connecting [latex](0, 2)[/latex] and [latex](4, 2)[/latex], and it can be parameterized by the equations [latex]x(t)=t[/latex], [latex]y(t)=2[/latex] for [latex]0\leq t\leq 4[/latex]. Again, define [latex]g(t)=f(x(t), y(t))[/latex]. This gives [latex]g(t)=t^{2}-8t+36[/latex]. The critical value corresponds to the point [latex](4, 2)[/latex]. So, calculating [latex]f(4, 2)[/latex] gives the [latex]z[/latex]-value [latex]20[/latex].

[latex]L_4[/latex] is the line segment connecting [latex](0, 0)[/latex] and [latex](0, 2)[/latex], and it can be parameterized by the equations [latex]x(t)=0[/latex], [latex]y(t)=t[/latex] for [latex]0\leq t\leq 2[/latex]. This time [latex]g(t)=4t^{2}-2t+24[/latex] and the critical value [latex]t=\frac{1}{4}[/latex]corresponds to the point [latex](0, \frac{1}{4})[/latex]. Calculating [latex]f(0, \frac{1}{4})[/latex] gives the [latex]z[/latex]-value [latex]23.75[/latex]. We also need to find the values of [latex]f(x, y)[/latex] at the corners of its domain. These corners are located at [latex](0, 0)[/latex], [latex](4, 0)[/latex], [latex](4, 2)[/latex], and [latex](0, 2)[/latex]:
[latex]\hspace{5cm}\begin{align} f(0,0)&=(0)^2-2(0)(0)+4(0)^2-4(0)-2(0)+24=24 \\ f(4,0)&=(4)^2-2(4)(0)+4(0)^2-4(4)-2(0)+24=24 \\ f(4,2)&=(4)^2-2(4)(2)+4(2)^2-4(4)-2(2)+24=20 \\ f(0,2)&=(0)^2-2(0)(2)+4(2)^2-4(0)-2(2)+24=36. \end{align}[/latex]

The absolute maximum value is [latex]36[/latex], which occurs at [latex](0, 2)[/latex], and the global minimum value is [latex]17[/latex], which occurs at [latex](3, 1)[/latex] as shown in the following figure.

The function f(x, y) = x2 − 2xy – 4x + 4y2 – 2y + 24 is shown with local minima at (4, 2, 20) and (2, 0, 20) and local maximum at (0, 2, 36). The shape is a plane curving up on the corners significantly near (0, 2) and slightly less near (4, 0).

Figure 2. The function [latex]\small{f(x,y)}[/latex] has one global minimum and one global maximum over its domain.

b. Using the problem-solving strategy, step 1 involves finding the critical points of [latex]g[/latex] on its domain. Therefore, we first calculate [latex]g_x(x, y)[/latex] and [latex]g_y(x, y)[/latex], then set them each equal to zero:

[latex]\hspace{8cm}\begin{align}g_x(x, y)&=2y+4 \\g_y(x, y)&=2y-6. \end{align}[/latex]

Setting them equal to zero yields the system of equations

[latex]\hspace{8cm}\begin{align} 2x+4&=0 \\ 2y-6&=0. \end{align}[/latex]
The solution to this system is [latex]x=-2[/latex] and [latex]y=3[/latex]. Therefore, [latex](-2, 3)[/latex] is a critical point of [latex]g[/latex]. Calculating [latex]g(-2, 3)[/latex], we get

[latex]g(-2,3)=(-2)^2+3^2+4(-2)-6(3)=4+9-8-18=-13.[/latex]

The next step involves finding the extrema of [latex]g[/latex] on the boundary of its domain. The boundary of its domain consists of a circle of radius [latex]4[/latex] centered at the origin as shown in the following graph.

A filled-in circle marked D of radius four with center at the origin.

Figure 3. Graph of the domain of the function [latex]\small{g(x,y)=x^{2}+y^{2}+4x-6y}[/latex].

The boundary of the domain of [latex]g[/latex] can be parameterized using the functions [latex]x(t)=4\cos t[/latex], [latex]y(t)=4\sin t[/latex] for [latex]0\leq{t}\leq2\pi[/latex]. Define [latex]h(t)=g(x(t), y(t))[/latex]:
[latex]\hspace{5cm}\begin{align} h(t)&=g(x(t),y(t)) \\ &=(4\cos{t})^2+(4\sin t)^2+4(4\cos{t})-6(4\sin t) \\ &=16\cos^2{t}+16\sin^2{t}+16\cos{t}-24\sin t \\ &=16+16\cos{t}-24\sin t.\end{align}[/latex]
Setting [latex]h'(t)=0[/latex] leads to
[latex]\hspace{5cm}\begin{align}-16\sin{t}-24\cos{t}&=0 \\-16\sin{t}&=24\cos{t} \\\frac{-16\sin{t}}{-16\cos{t}}&=\frac{24\cos{t}}{-16\cos{t}} \\\tan{t}&=-\frac32\end{align}[/latex]
This equation has two solutions over the interval [latex]0\leq{t}\leq2\pi[/latex]. One is [latex]t=\pi-\arctan{(\frac32)}[/latex] and the other is [latex]t=2\pi-\arctan{(\frac32)}[/latex]. For the first angle,
[latex]\hspace{4cm}\begin{align}\sin{t}&=\sin{(\pi-\arctan{(\frac32)})}=\sin{(\arctan{(\frac32)})}=\frac{3\sqrt{13}}{13} \\\cos{t}&=\cos{(\pi-\arctan{(\frac32)})}=-\cos{(\arctan{(\frac32)})}=-\frac{2\sqrt{13}}{13}\end{align}[/latex]
Therefore, [latex]x(t)=4\cos{t}=\frac{8\sqrt{13}}{13}[/latex] and [latex]y(t)=4\sin{t}=-\frac{12\sqrt{13}}{13}[/latex], so [latex]\left(\frac{8\sqrt{13}}{13},-\frac{12\sqrt{13}}{13}\right)[/latex] is a critical point on the boundary and[latex]\hspace{4cm}\begin{align}g\left(-\frac{8\sqrt{13}}{13},\frac{12\sqrt{13}}{13}\right)&=\left(-\frac{8\sqrt{13}}{13}\right)^2+\left(\frac{12\sqrt{13}}{13}\right)^2+4\left(-\frac{8\sqrt{13}}{13}\right)-6\left(\frac{12\sqrt{13}}{13}\right) \\&=\frac{144}{13}+\frac{64}{13}-\frac{32\sqrt{13}}{13}-\frac{72\sqrt{13}}{13} \\&=\frac{208-104\sqrt{13}}{13}\approx-12.844.\end{align}[/latex]
For the second angle,
[latex]\hspace{5cm}\begin{align}\sin{t}&=\sin{(2\pi-\arctan{(\frac32)})}=-\sin{(\arctan{(\frac32)})}=-\frac{3\sqrt{13}}{13} \\\cos{t}&=\cos{(2\pi-\arctan{(\frac32)})}=\cos{(\arctan{(\frac32)})}=\frac{2\sqrt{13}}{13}.\end{align}[/latex]
Therefore, [latex]x(t)=4\cos{t}=\frac{8\sqrt{13}}{13}[/latex] and [latex]y(t)=4\sin{t}=-\frac{12\sqrt{13}}{13}[/latex], so [latex]\left(\frac{8\sqrt{13}}{13},-\frac{12\sqrt{13}}{13}\right)[/latex] is a critical point on the boundary and[latex]\hspace{4cm}\begin{align}g\left(\frac{8\sqrt{13}}{13},-\frac{12\sqrt{13}}{13}\right)&=\left(\frac{8\sqrt{13}}{13}\right)^2+\left(-\frac{12\sqrt{13}}{13}\right)^2+4\left(\frac{8\sqrt{13}}{13}\right)-6\left(-\frac{12\sqrt{13}}{13}\right) \\&=\frac{144}{13}+\frac{64}{13}+\frac{32\sqrt{13}}{13}+\frac{72\sqrt{13}}{13} \\&=\frac{208+104\sqrt{13}}{13}\approx44.844.\end{align}[/latex]The absolute minimum of [latex]g[/latex] is [latex]-13[/latex] which is attained at the point [latex](-2, 3)[/latex], which is an interior point of [latex]D[/latex]. The absolute maximum of [latex]g[/latex] is approximately equal to [latex]44.844[/latex], which is attained at the boundary point [latex]\left(\frac{8\sqrt{13}}{13},-\frac{12\sqrt{13}}{13}\right)[/latex]. These are the absolute extrema of [latex]g[/latex] on [latex]D[/latex] as shown in the following figure.

The function f(x, y) = x2 + 4x + y2 – 6y is shown with local minimum at (–12/5, 16/5, –64/5) and local maximum at (12/5, −16/5, 224/5). The shape is a plane curving up from near (−4, 4) to (4, −4).

Figure 4. The function [latex]\small{f(x,y)}[/latex] has a local minimum and a local maximum.

Try it

Use the problem-solving strategy for finding absolute extrema of a function to find the absolute extrema of the function

[latex]f(x, y)=4x^{2}-2xy+6y^{2}-8x+2y+3[/latex]

on the domain defined by [latex]0\leq x\leq 2[/latex] and [latex]-1\leq y\leq 3[/latex].

Show Solution

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

You can view the transcript for “CP 4.36” here (opens in new window).

Module 4: Differentiation of Functions of Several Variables