Worked examples · Sheet 2

Absolute maxima and minima on closed and bounded sets

Theory used: The method for absolute extrema Fermat's theorem in two variables Stationary point

Exercise 1

Find the absolute maximum and minimum of $f (x, y) = x^{2} + y^{2} - 2 x$ on the rectangle $R = [0, 3] \times [- 1, 1]$ .

{f_{x} = 2 x - 2 = 0 f_{y} = 2 y = 0 ⟺ (x, y) = (1, 0), f (1, 0) = - 1

y = \pm 1, x \in [0, 3] : x = 0, y \in [- 1, 1] : x = 3, y \in [- 1, 1] : f = x^{2} - 2 x + 1 = (x - 1)^{2} \in [0, 4] f = y^{2} \in [0, 1] f = 3 + y^{2} \in [3, 4]

R max f = f (3, \pm 1) = 4, R min f = f (1, 0) = - 1

The rectangle is closed and bounded and $f$ is a polynomial, hence continuous; by the Weierstrass theorem the absolute maximum and minimum exist. First we look for the interior stationary points. The system gives the single point $(1, 0)$ , which is interior to $R$ because $0 < 1 < 3$ and $- 1 < 0 < 1$ .

The boundary consists of four sides. On the horizontal sides $y = \pm 1$ the function becomes $(x - 1)^{2}$ with $x \in [0, 3]$ , which goes from $0$ , at $x = 1$ , to $4$ , at $x = 3$ . On the side $x = 0$ we have $y^{2}$ , between $0$ and $1$ ; on the side $x = 3$ we have $3 + y^{2}$ , between $3$ and $4$ . Comparing all the values found, the largest is $4$ , at the vertices $(3, 1)$ and $(3, - 1)$ , and the smallest is $- 1$ , at the interior point $(1, 0)$ . Hence the absolute maximum is $4$ and the absolute minimum is $- 1$ .

Exercise 2

Find the absolute maximum and minimum of $f (x, y) = x^{2} - x y + y^{2} - x$ on the triangle $T$ with vertices $(0, 0)$ , $(2, 0)$ and $(0, 2)$ .

{f_{x} = 2 x - y - 1 = 0 f_{y} = - x + 2 y = 0 ⟺ {x = 2 y 3 y - 1 = 0 ⟺ (x, y) = (\frac{2}{3}, \frac{1}{3}), f (\frac{2}{3}, \frac{1}{3}) = - \frac{1}{3}

y = 0, x \in [0, 2] : x = 0, y \in [0, 2] : y = 2 - x, x \in [0, 2] : φ_{1} (x) = x^{2} - x φ_{2} (y) = y^{2} φ_{3} (x) = 3 x^{2} - 7 x + 4

φ_{1}^{'} (x) = 2 x - 1 = 0 ⟺ x = \frac{1}{2}, f (\frac{1}{2}, 0) = - \frac{1}{4}

φ_{3}^{'} (x) = 6 x - 7 = 0 ⟺ x = \frac{7}{6}, f (\frac{7}{6}, \frac{5}{6}) = - \frac{1}{12}

f (0, 0) = 0, f (2, 0) = 2, f (0, 2) = 4

T max f = f (0, 2) = 4, T min f = f (\frac{2}{3}, \frac{1}{3}) = - \frac{1}{3}

The triangle

T

with the interior stationary point

(\frac{2}{3}, \frac{1}{3})

, the candidates on the sides and the vertex

(0, 2)

, where

f

is maximum

The triangle is the set $x \geq 0$ , $y \geq 0$ , $x + y \leq 2$ , closed and bounded, and $f$ is continuous, so the absolute extrema exist by the Weierstrass theorem. The first-order system gives the point $(\frac{2}{3}, \frac{1}{3})$ , which is interior because the coordinates are positive and their sum is $1 < 2$ .

On each side the equation of the side is substituted into $f$ . On the side $y = 0$ one obtains the parabola $x^{2} - x$ , with vertex at $x = \frac{1}{2}$ . On the side $x = 0$ one obtains $y^{2}$ , increasing on $[0, 2]$ , which gives no new points besides the vertices. On the hypotenuse one substitutes $y = 2 - x$ :

x^{2} - x (2 - x) + (2 - x)^{2} - x = x^{2} - 2 x + x^{2} + 4 - 4 x + x^{2} - x = 3 x^{2} - 7 x + 4,

with vertex at $x = \frac{7}{6}$ , that is, at the point $(\frac{7}{6}, \frac{5}{6})$ . The three vertices are added to the candidates. The smallest value is $- \frac{1}{3}$ , at the interior point, and the largest is $4$ , at the vertex $(0, 2)$ ; therefore the absolute minimum is $- \frac{1}{3}$ and the absolute maximum is $4$ .

Exercise 3

Find the absolute maximum and minimum of $f (x, y) = x^{2} + 2 y^{2} - x$ on the disk $D = {(x, y) : x^{2} + y^{2} \leq 1}$ .

{f_{x} = 2 x - 1 = 0 f_{y} = 4 y = 0 ⟺ (x, y) = (\frac{1}{2}, 0), f (\frac{1}{2}, 0) = - \frac{1}{4}

x^{2} + y^{2} = 1 ⟹ y^{2} = 1 - x^{2}, φ (x) = x^{2} + 2 (1 - x^{2}) - x = 2 - x^{2} - x, x \in [- 1, 1]

φ^{'} (x) = - 2 x - 1 = 0 ⟺ x = - \frac{1}{2}, φ (- \frac{1}{2}) = \frac{9}{4}, φ (1) = 0, φ (- 1) = 2

D max f = f (- \frac{1}{2}, \pm \frac{3}{2}) = \frac{9}{4}, D min f = f (\frac{1}{2}, 0) = - \frac{1}{4}

The closed disk is closed and bounded and $f$ is continuous, so the absolute maximum and minimum exist. The only stationary point is $(\frac{1}{2}, 0)$ , interior to the disk because $\frac{1}{4} < 1$ .

The boundary is the circle $x^{2} + y^{2} = 1$ . Only $y^{2}$ appears in the function, therefore it is convenient to substitute $y^{2} = 1 - x^{2}$ , and on the circle $f$ becomes the function of one variable $φ (x) = 2 - x^{2} - x$ , with $x \in [- 1, 1]$ . Its derivative vanishes at $x = - \frac{1}{2}$ , which corresponds to the two points $(- \frac{1}{2}, \pm \frac{3}{2})$ of the circle; the endpoints $x = \pm 1$ correspond to the points $(1, 0)$ and $(- 1, 0)$ . Comparing the values $- \frac{1}{4}$ , $\frac{9}{4}$ , $0$ and $2$ gives the absolute maximum $\frac{9}{4}$ and the absolute minimum $- \frac{1}{4}$ .