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Section 8.4 Absolute Maximum and Absolute Minimum

In the next example, we will see how to find absolute extrema of a function \(f(x,y)\) over a closed bounded region \(R\text{.}\)

Example 8.17. Absolute Extrema over a Rectangle.

Find the absolute maximum and minimum values of the function \(f(x,y) = x^2 - 2xy + 2y\) on the rectangle
\begin{equation} R = \{(x,y) : 0 \le x \le 3,\; 0 \le y \le 2\}.\tag{8.15} \end{equation}

Solution.

Step (I): We begin by finding the critical points of \(f(x,y)\) that are also in the set \(R\text{:}\)
\begin{align*} f_x \amp= 2x - 2y = 0 \Rightarrow x = y\\ f_y \amp= -2x + 2 = 0 \Rightarrow x = 1. \end{align*}
Therefore, the only critical point is \((1,1)\text{,}\) with \(f(1,1) = 1\text{.}\)
Step (II): We find the extreme values of \(f(x,y)\) on the boundary of \(R\text{,}\) i.e. along the four edges of the rectangle denoted by \(L_1\text{,}\) \(L_2\text{,}\) \(L_3\text{,}\) and \(L_4\text{,}\) shown in Figure 8.19.
  1. Along the first edge \(L_1\text{,}\) we have \(f(x,0) = x^2\) and since \(0 \le x \le 3\text{,}\) the minimum value is \(f(0,0) = 0\) and the maximum value is \(f(3,0) = 9\text{.}\)
  2. Along the second edge \(L_2\text{,}\) we have \(f(3,y) = 9 - 4y\) and since \(0 \le y \le 2\text{,}\) the minimum value is \(f(3,2) = 1\) and the maximum value is \(f(3,0) = 9\text{.}\)
  3. Along the third edge \(L_3\text{,}\) we have \(f(x,2) = x^2 - 4x + 4 = (x-2)^2\) and since \(0 \le x \le 3\text{,}\) the minimum value is \(f(2,2) = 0\) and the maximum value is \(f(0,2) = 4\text{.}\)
  4. Along the fourth edge \(L_4\text{,}\) we have \(f(0,y) = 2y\) and since \(0 \le y \le 2\text{,}\) the minimum value is \(f(0,0) = 0\) and the maximum value is \(f(0,2) = 4\text{.}\)
Step (III): We compare the values obtained in the previous steps. The absolute maximum value of \(f(x,y)\) on the rectangle \(R\) is \(f(3,0) = 9\) and its absolute minimum is \(f(0,0) = f(2,2) = 0\text{.}\) The candidate points and the surface are shown in Figure 8.20 and Figure 8.21.
Figure 8.18. The closed bounded rectangle \(R = \{(x,y) : 0 \le x \le 3,\; 0 \le y \le 2\}\) with its four edges \(L_1\text{,}\) \(L_2\text{,}\) \(L_3\text{,}\) and \(L_4\text{,}\) and the critical point \((1,1)\) in its interior.
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Figure 8.19. The rectangle \(R\) and its four edges. To find the absolute extrema, we evaluate \(f\) at the critical points inside \(R\) and compare with the extreme values of \(f\) along each edge.
Figure 8.20. The surface \(z = x^2 - 2xy + 2y\) over the rectangle \(R\text{.}\) The absolute maximum \(f(3,0) = 9\) and the absolute minimum \(f(0,0) = f(2,2) = 0\) occur on the boundary of \(R\text{.}\)
Diagram Exploration Keyboard Controls
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Enter, A Activate keyboard driven exploration
B Activate menu driven exploration
Escape Leave exploration mode
Cursor down Explore next lower level
Cursor up Explore next upper level
Cursor right Explore next element on level
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X Toggle expert mode
W Extra details if available
Space Repeat speech
M Activate step magnification
Comma Activate direct magnification
N Deactivate magnification
Z Toggle subtitles
C Cycle contrast settings
T Monochrome colours
L Toggle language (if available)
K Kill current sound
Y Stop sound output
O Start and stop sonification
P Repeat sonification output
Figure 8.21. The candidate points for the absolute extrema of \(f(x,y) = x^2 - 2xy + 2y\) on \(R\text{:}\) the interior critical point and the extreme points found along the four edges. Comparing the values of \(f\text{,}\) the absolute maximum is \(f(3,0) = 9\) and the absolute minimum is \(f(0,0) = f(2,2) = 0\text{.}\)
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