1.7 Absolute Maximum and Minimum Values
For a function of one variable, a continuous function on a closed interval attains a maximum and a minimum. The analogue in the plane replaces the closed interval by a set that is closed (it contains all its boundary points) and bounded (it lies inside some disc).
Theorem 1.7.1 (Extreme value theorem). If \(f\) is continuous on a closed bounded set \(D\) in \(\mathbb {R}^2\), then \(f\) attains an absolute maximum value \(f(x_1,y_1)\) and an absolute minimum value \(f(x_2,y_2)\) at some points \((x_1,y_1)\) and \((x_2,y_2)\) of \(D\).
An extreme value attained at an interior point of \(D\) is also a local extreme value, so by Theorem 1.5.2 it occurs at a critical point. Every other extreme value is attained on the boundary. This gives a procedure.
Closed Interval Method
To find the absolute maximum and minimum of a continuous function \(f\) on a closed bounded set \(D\):
- 1.
- Find the values of \(f\) at the critical points of \(f\) in the interior of \(D\).
- 2.
- Find the extreme values of \(f\) on the boundary of \(D\).
- 3.
- The largest of the values from steps 1 and 2 is the absolute maximum value, and the smallest is the absolute minimum value.
Step 2 usually reduces to one-variable problems, one for each piece of the boundary, or is done with Lagrange multipliers.
Example 1.7.2. Find the absolute maximum and minimum values of \(f(x,y) = x^2 - 2xy + 2y\) on the rectangle \(D = \{(x,y) \mid 0\leq x \leq 3,\ 0 \leq y \leq 2 \}\).
Solution. The polynomial \(f\) is continuous on the closed bounded rectangle, so the extreme values exist.
Interior. \(f_x = 2x -2y\) and \(f_y = -2x + 2\) vanish only at \((1,1)\), which lies inside \(D\), and \(f(1,1) = 1\).
Boundary. The boundary consists of the four segments \(L_1\) to \(L_4\).
- \(-\)
- On \(L_1\): \(y = 0\), \(0\leq x\leq 3\), and \(f = x^2\), which increases from \(0\) to \(9\).
- \(-\)
- On \(L_2\): \(x = 3\), \(0\leq y\leq 2\), and \(f = 9 - 4y\), which decreases from \(9\) to \(1\).
- \(-\)
- On \(L_3\): \(y = 2\), \(0\leq x\leq 3\), and \(f = x^2 - 4x + 4 = (x-2)^2\), with minimum \(0\) at \(x = 2\) and maximum \(4\) at \(x = 0\).
- \(-\)
- On \(L_4\): \(x = 0\), \(0\leq y\leq 2\), and \(f = 2y\), from \(0\) to \(4\).
Comparison. The candidates are \(1\) from the interior and \(0\), \(1\), \(4\), \(9\) from the boundary. The absolute maximum is \(f(3,0) = 9\), and the absolute minimum is \(0\), attained at \((0,0)\) and at \((2,2)\).
Example 1.7.3. Find the absolute extreme values of \(f(x,y) = x^2 + y^2 - 2x - 2y\) on the closed triangle with vertices \((0,0)\), \((4,0)\) and \((0,4)\).
Solution. Interior. \(f_x = 2x - 2 = 0\) and \(f_y = 2y - 2 = 0\) give \((1,1)\), inside the triangle, with \(f(1,1) = -2\).
Boundary. On \(y = 0\), \(0\leq x\leq 4\): \(f = x^2 - 2x\), with minimum \(-1\) at \(x = 1\) and values \(0\) and \(8\) at the ends. On \(x = 0\) the same by symmetry. On the hypotenuse \(y = 4 - x\), \(0\leq x\leq 4\): \[f = x^2 + (4-x)^2 - 2x - 2(4-x) = 2x^2 - 8x + 8 = 2(x-2)^2,\] with minimum \(0\) at \((2,2)\) and maximum \(8\) at the ends.
Comparison. The absolute minimum is \(-2\) at \((1,1)\), and the absolute maximum is \(8\), at \((4,0)\) and \((0,4)\).
Example 1.7.4. Find the absolute extreme values of \(f(x,y) = 2x^2 + y^2 - 4x + 1\) on the disc \(x^2 + y^2\leq 4\), using a Lagrange multiplier on the boundary.
Solution. Interior. \(f_x = 4x - 4\) and \(f_y = 2y\) vanish at \((1,0)\), inside the disc, where \(f = -1\).
Boundary. On \(x^2 + y^2 = 4\), the equations \(\nabla f = \lambda \nabla g\) are \(4x - 4 = 2\lambda x\) and \(2y = 2\lambda y\). The second gives \(y = 0\) or \(\lambda = 1\). If \(y = 0\), then \(x = \pm 2\), with \(f(2,0) = 1\) and \(f(-2,0) = 17\). If \(\lambda = 1\), the first gives \(4x - 4 = 2x\), so \(x = 2\) and \(y = 0\) again.
Comparison. The absolute maximum is \(17\), at \((-2,0)\), and the absolute minimum is \(-1\), at \((1,0)\).
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