Worked examples · Sheet 1

Stationary points and the Hessian test

Theory used: Stationary point Fermat's theorem in two variables Hessian test Saddle point

Exercise 1

Find and classify the stationary points of

f(x, y) = x^2 + xy + y^2 - 3x .

\begin{cases} f_x = 2x + y - 3 = 0 \\[4pt] f_y = x + 2y = 0 \end{cases} \iff \begin{cases} x = -2y \\[4pt] -4y + y - 3 = 0 \end{cases} \iff \begin{cases} x = 2 \\[4pt] y = -1 \end{cases}

f_{xx} = 2, \qquad f_{xy} = 1, \qquad f_{yy} = 2

H(2, -1) = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}, \qquad \det H(2, -1) = 4 - 1 = 3 > 0, \qquad f_{xx} = 2 > 0

f(2, -1) = 4 - 2 + 1 - 6 = -3

The function is a polynomial, so it has partial derivatives at every point of the plane, and by Fermat’s theorem in two variables its relative extrema are among the points where the gradient is zero. From the second equation one obtains $x = -2y$ ; substituting into the first gives $-3y - 3 = 0$ , that is, $y = -1$ , and hence $x = 2$ . The only stationary point is $(2, -1)$ .

The second derivatives are constant, therefore the Hessian is the same at every point. Its determinant is positive and $f_{xx}$ is positive, so by the Hessian test the point $(2, -1)$ is a relative minimum point, with value $f(2, -1) = -3$ .

Exercise 2

Find and classify the stationary points of

f(x, y) = x^3 - 3x + y^2 - 2y .

\begin{cases} f_x = 3x^2 - 3 = 0 \\[4pt] f_y = 2y - 2 = 0 \end{cases} \iff \begin{cases} x = \pm 1 \\[4pt] y = 1 \end{cases}

f_{xx} = 6x, \qquad f_{xy} = 0, \qquad f_{yy} = 2

H(1, 1) = \begin{pmatrix} 6 & 0 \\ 0 & 2 \end{pmatrix}, \qquad \det H(1, 1) = 12 > 0, \qquad f_{xx}(1, 1) = 6 > 0

H(-1, 1) = \begin{pmatrix} -6 & 0 \\ 0 & 2 \end{pmatrix}, \qquad \det H(-1, 1) = -12 < 0

f(1, 1) = 1 - 3 + 1 - 2 = -3, \qquad f(-1, 1) = -1 + 3 + 1 - 2 = 1

The two equations of the system each contain only one variable, so they are solved separately. The first gives $x^2 = 1$ , that is, $x = 1$ or $x = -1$ ; the second gives $y = 1$ . The stationary points are therefore $(1, 1)$ and $(-1, 1)$ .

The derivative $f_{xx} = 6x$ depends on the point, so the Hessian must be computed at each one. At $(1, 1)$ the determinant is positive and $f_{xx} > 0$ , therefore the point is a relative minimum point, with value $-3$ . At $(-1, 1)$ the determinant is negative, therefore the point is a saddle point. As a check, along the line $y = 1$ the function equals $x^3 - 3x - 1$ , which has a maximum at $x = -1$ , while along the line $x = -1$ it equals $y^2 - 2y + 2$ , which has a minimum at $y = 1$ .

Exercise 3

Find and classify the stationary points of

f(x, y) = x^2 y - 2xy + y^2 .

\begin{cases} f_x = 2xy - 2y = 2y\,(x - 1) = 0 \\[4pt] f_y = x^2 - 2x + 2y = 0 \end{cases}

y = 0: \quad x^2 - 2x = 0 \iff x = 0 \ \text{ or } \ x = 2

x = 1: \quad 1 - 2 + 2y = 0 \iff y = \frac{1}{2}

f_{xx} = 2y, \qquad f_{xy} = 2x - 2, \qquad f_{yy} = 2

\begin{aligned} H(0, 0) &= \begin{pmatrix} 0 & -2 \\ -2 & 2 \end{pmatrix}, &\quad \det H(0, 0) &= -4 < 0 \\[6pt] H(2, 0) &= \begin{pmatrix} 0 & 2 \\ 2 & 2 \end{pmatrix}, &\quad \det H(2, 0) &= -4 < 0 \\[6pt] H\left( 1, \frac{1}{2} \right) &= \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}, &\quad \det H\left( 1, \frac{1}{2} \right) &= 2 > 0 \end{aligned}

f\left( 1, \frac{1}{2} \right) = \frac{1}{2} - 1 + \frac{1}{4} = -\frac{1}{4}

The first equation is a product equal to zero, so it splits into two cases, $y = 0$ or $x = 1$ , and each case is substituted into the second equation. With $y = 0$ the second becomes $x^2 - 2x = x\,(x - 2) = 0$ , which gives $x = 0$ and $x = 2$ . With $x = 1$ the second becomes $2y - 1 = 0$ , which gives $y = \dfrac{1}{2}$ . There are therefore three stationary points, $(0, 0)$ , $(2, 0)$ and $\left( 1, \dfrac{1}{2} \right)$ .

At the points $(0, 0)$ and $(2, 0)$ the determinant of the Hessian is negative, therefore by the Hessian test they are two saddle points. At the point $\left( 1, \dfrac{1}{2} \right)$ the determinant is positive and $f_{xx} = 1 > 0$ , therefore the point is a relative minimum point, with value $-\dfrac{1}{4}$ . The minimum is not absolute, because along the line $x = 3$ the function equals $3y + y^2$ , and at $y = -\dfrac{3}{2}$ this value is $-\dfrac{9}{4}$ , smaller than $-\dfrac{1}{4}$ .