Title: Default for 18.3 Stationary Points
Course: HELM Workbooks 2023-2024

The calculation of the optimum value of a function of two variables is a common requirement in many areas of engineering, for example in thermodynamics. Unlike the case of a function of one variable we have to use more complicated criteria to distinguish between the various types of stationary point.

The stationary points of a function of two variables

The Figure below shows a computer generated picture of the surface defined by the function \(z=x^3+y^3-3x-3y,\) where both \(x\) and \(y\) take values in the interval \([-1.8,1.8]\).

[Image: The plot of the surface z = x^3 + y^3 - 3x - 3y in (x, y, z) coordinates]

There are four features of particular interest on the surface. At point \(A\) there is a local maximum, at \(B\) there is a local minimum, and at \(C\) and \(D\) there are what are known as saddle points. At \(A\) the surface is at its greatest height in the immediate neighbourhood. If we move on the surface from \(A\) we immediately loose height no matter in which direction we travel. At \(B\) the surface is at its least height in the neighbourhood. If we move on the surface from \(B\) we immediately gain height, no matter in which direction we travel. The features at \(C\) and \(D\) are quite different. In some directions as we move away from these points along the surface we lose height whilst in others we gain height. The similarity in shape to a horse’s saddle is evident. At each point \(P\) of a smooth surface one can draw a unique plane which touches the surface there. This plane is called the tangent plane at \(P\). (The tangent plane is a natural generalisation of the tangent line which can be drawn at each point of a smooth curve.) In the Figure above, at each of the points \(A,B,C,D\) the tangent plane to the surface is horizontal at the point of interest. Such points are thus known as stationary points of the function. In the next subsections we show how to locate stationary points and how to determine their nature using partial differentiation of the function \(f(x,y)\).

Q: Exercise

Given that \(f(x) = [value]\), identify the stationary point for this function.

Classify the stationary point as either a local minimum, local maximum or a saddle point.

\(\dfrac{\partial f}{\partial x} = \dfrac{\partial f}{\partial y} = 0\) at \((x, y)\) =

This point is a:
Feedback: We are asked to find the stationary points of the function \( f(x,y) = [value].\) First, find the partial derivatives of \(f(x,y):\) \(f_x=[value] \\ f_y=[value]\) Now we solve the equations \(f_x = 0\) and \(f_y= 0:\) \(f_x = 0 \Rightarrow [value] = 0 .\) \(f_y = 0 \Rightarrow [value] = 0 .\) Solving the system of equations we find that the stationary point is \( (-1, 3) .\) Then we calculate \( f_{xx}, \, f_{yy}, \, f_{xy} \, \text{and} \, D:\) \( f_{xx} = [value] \) \( f_{yy} = [value] \) \( f_{xy} = [value] \) \( D = f_{xx}f_{yy}- (f_{xy})^2 = [value] \) We have \(D<0 \) so the stationary point is a saddle point.
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Location of stationary points

As we said in the previous subsection, the tangent plane to the surface \(z=f(x,y)\) is horizontal at a stationary point. A condition which guarantees that the function \(f(x,y)\) will have a stationary point at a point \((x_0,y_0)\) is that, at that point both \(f_x=0\) and \(f_y=0\) simultaneously.

The nature of a stationary point

We state, without proof, a relatively simple test to determine the nature of a stationary point, once located. If the surface is very flat near the stationary point then the test will not be sensitive enough to determine the nature of the point. The test is dependent upon the values of the second order derivatives: \(f_{xx}, f_{yy}, f_{xy}\) and also upon a combination of second order derivatives denoted by \(D\) where \( D \equiv\dfrac{\partial^2f}{\partial x^2}\times\dfrac{\partial^2f}{\partial y^2}-\left(\dfrac{\partial^2f}{\partial x\partial y}\right)^2\) , which is also expressible as \(D\equiv f_{xx}f_{yy}- (f_{yx})^2\). The test is as follows:

Key Point

Test to Determine the Nature of Stationary Points

1. At each stationary point work out the three second order partial derivatives.

2. Calculate the value of \(D=f_{xx}f_{yy}-(f_{yx})^2\) at each stationary point.Then, test each stationary point in turn:

3. If \(D<0\) the stationary point is a saddle point.

If \(D>0\) and \(\dfrac{\partial^2f}{\partial x^2}>0\) the stationary point is a local minimum.

If \(D>0\) and \(\dfrac{\partial^2f}{\partial x^2}<0\) the stationary point is a local maximum.

If \(D=0\) then the test is inconclusive (we need an alternative test).

Example

The function: \(f(x,y)=x^4+y^4-36 xy\) has stationary points at \((0,0),\ (-3,-3),\ (3,3).\) Use Key Point 4 to determine the nature of each stationary point.

Solution

We have \(\dfrac{\partial f}{\partial x}=f_x=4x^3-36y\) and \(\dfrac{\partial f}{\partial y}=f_y=4y^3-36x\).

Then \(\dfrac{\partial^2f}{\partial x^2}=f_{xx}=12x^2, \dfrac{\partial^2f}{\partial y^2}=f_{yy}=12y^2\), \(\dfrac{\partial^2f}{\partial x\partial y}=f_{yx}=-36\).

A tabular presentation is useful for calculating \(D=f_{xx}f_{yy}-(f_{yx})^2\):

 | Point | Point | Point
Derivatives | (0, 0) | (-3, -3) | (3, 3)
\(f_{xx}\) | 0 | 108 | 108
\(f_{yy}\) | 0 | 108 | 108
\(f_{xy}\) | -36 | -36 | -36
D | <0 | >0 | >0

\((0,0)\) is a saddle point; \((-3,-3)\) and \((3,3)\) are both local minima.

Exercise

If \(f(x,y) = [value]\) find the stationary points for this function

For each stationary point classify it as either a local minimum, a local maximum or a saddle point.

1. \(\dfrac{\partial f}{\partial x} = \dfrac{\partial f}{\partial y} = 0\) at \((x, y)\) =
This point is a:
2. \(\dfrac{\partial f}{\partial x} = \dfrac{\partial f}{\partial y} = 0\) at \((x, y)\) =
This point is a:
Feedback: We are asked to find the stationary points of the function \( f(x,y) = [value].\) First, find the partial derivatives of \(f(x,y):\) \(f_x=[value] \\ f_y=[value]\) Now we solve the equations \(f_x = 0\) and \(f_y= 0:\) \(f_x = 0 \Rightarrow [value] = 0 \, \Rightarrow x= [value]\text{ or } x= [value].\) \(f_y = 0 \Rightarrow [value] = 0 \, \Rightarrow y= [value].\) Hence the stationary points are \( ([value], [value]) \) and \( ([value], [value]) .\) Then we calculate \( f_{xx}, \, f_{yy}, \, f_{xy} \,\, \text{and} \,\, D:\) \( f_{xx} = [value] \) \( f_{yy} = [value] \) \( f_{xy} = [value] \) \( D = f_{xx}f_{yy}- (f_{xy})^2 = [value] \) Now complete the table below and determine the nature of the stationary points: | Point | Point Derivatives | ([value], [value]) | ([value], [value]) \(f_{xx}\) | [value] | [value] \(f_{yy}\) | [value] | [value] \(f_{xy}\) | [value] | [value] \(D\) | \([value]<0\) | \([value]>0\) Hence, \(([value], [value])\) is a local maximum and \(([value], [value])\) is a saddle point.

Q: Exercise

\(f(x,y) = [value]\), identify the stationary point for this function.

Classify the stationary point as either a local minimum, a local maximum or a saddle point.

\(\dfrac{\partial f}{\partial x} = \dfrac{\partial f}{\partial y} = 0\) at \((x, y)\) =

This point is a:
Feedback: We are asked to find the stationary points of the function \( f(x,y) = [value].\) First, find the partial derivatives of \(f(x,y):\) \(f_x=[value] \\ f_y=[value]\) Now we solve the equations \(f_x = 0\) and \(f_y= 0:\) \(f_x = 0 \Rightarrow [value] = 0 \, \Rightarrow x= [value] .\) \(f_y = 0 \Rightarrow [value] = 0 \, \Rightarrow y= [value] .\) These equations have solution \( x=y= 0,\) hence the stationary point is \( (0, 0) .\) Then we calculate \( f_{xx}, \, f_{yy}, \, f_{xy} \, \text{ and } \, D:\) \( f_{xx} = [value] \) \( f_{yy} = [value] \) \( f_{xy} = [value] \) \( D = f_{xx}f_{yy}- (f_{xy})^2 = [value] \) We have \(D<0 \) and \(f_{xx} <0\) so the stationary point is a saddle point.
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Exercise

Find the stationary points for the function \(f(x,y) = [value]\) .

For each stationary point classify it as either a local minimum, a local maximum or a saddle point.

1. \(\dfrac{\partial f}{\partial x} = \dfrac{\partial f}{\partial y} = 0\) at \((x, y)\) =
This point is a:
2. \(\dfrac{\partial f}{\partial x} = \dfrac{\partial f}{\partial y} = 0\) at \((x, y)\) =
This point is a:
Feedback: We are asked to find the stationary points of the function \( f(x,y) = [value].\) First, find the partial derivatives of \(f(x,y):\) \(f_x=[value] \\ f_y=[value]\) Now we solve the equations \(f_x = 0\) and \(f_y= 0:\) \(f_x = 0 \Rightarrow [value] = 0 \, \Rightarrow x= [value] .\) \(f_y = 0 \Rightarrow [value] = 0 \, \Rightarrow y= [value] \text{ or } y= [value].\) Hence the stationary points are \( ([value], [value]) \) and \( ([value], [value]) .\) Then we calculate \( f_{xx}, \, f_{yy}, \, f_{xy} \, \text{and} \, D:\) \( f_{xx} = [value] \) \( f_{yy} = [value] \) \( f_{xy} = [value] \) \( D = f_{xx}f_{yy}- (f_{xy})^2 = [value] \) Now complete the table below and determine the nature of the stationary points: | Point | Point Derivatives | ([value], [value]) | ([value], [value]) \(f_{xx}\) | [value] | [value] \(f_{yy}\) | [value] | [value] \(f_{xy}\) | [value] | [value] \(D\) | \([value]>0\) | \([value]<0\) Hence, \(([value], [value])\) is a local minimum and \(([value], [value])\) is a saddle point.

Exercise

Find the stationary points for the function \(f(x,y) = [value]\).

For each stationary point classify it as either a local minimum, a local maximum or a saddle point.

1. \(\dfrac{\partial f}{\partial x} = \dfrac{\partial f}{\partial y} = 0\) at \((x, y)\) =
This point is a:
2. \(\dfrac{\partial f}{\partial x} = \dfrac{\partial f}{\partial y} = 0\) at \((x, y)\) =
This point is a:
3. \(\dfrac{\partial f}{\partial x} = \dfrac{\partial f}{\partial y} = 0\) at \((x, y)\) =
This point is a:
Feedback: We are asked to find the stationary points of the function \( f(x,y) = [value] .\) \(f_x=[value] \\ f_y=[value]\) Now we solve the equations \(f_x = 0\) and \(f_y= 0:\) \[ f_x = 0 \Rightarrow [value] = [value] =0 \Rightarrow\] \[ x=0 \text{ or } (x^2+y^2-1)=0. \] and \[f_y = 0 \Rightarrow [value] = [value] =0 \Rightarrow\] \[y=0 \text{ or } (x^2+y^2-1)=0. \] Solving the equation \( x^2+y^2-1 = 0\) gives the extra set of integer solutions \( (1,0), \, (-1,0) .\) Hence the stationary points are \( (0,0), \, (1,0) \) and \( (-1,0). \) Then we calculate \( f_{xx}, \, f_{yy}, \, f_{xy} \, \text{and} \, D:\) \( f_{xx} = [value] = [value] \) \( f_{yy} = [value] = [value] \) \( f_{xy} = [value] \) \( D = f_{xx}f_{yy}- (f_{xy})^2 = [value] \) Now complete the table below and determine the nature of the stationary points: | Point | Point | Point Derivatives | \((0,0)\) | \((1,0)\) | \((-1,0)\) \(f_{xx}\) | [value] | [value] | [value] \(f_{yy}\) | [value] | [value] | [value] \(f_{xy}\) | [value] | [value] | [value] \(D\) | \([value]<0\) | \([value]>0\) | \([value]>0\) Hence, \((0,0)\) is a saddle point and \((1,0), \, (-1,0)\) are local minima.

Exercise

Identify the coordinate of the stationary point for the function \(f(x) = [value]\) .

Classify this stationary point as either a local minimum, a local maximum or a saddle point.

\(\dfrac{\partial f}{\partial x} = \dfrac{\partial f}{\partial y} = 0\) at \((x, y)\) =

This point is a:
Feedback: We are asked to find the stationary points of the function \( f(x,y) = [value] = [value].\) First, find the partial derivatives of \(f(x,y):\) \(f_x=[value] \\ f_y=[value]\) Now we solve the equations \(f_x = 0\) and \(f_y= 0:\) \(f_x = 0 \Rightarrow [value] = 0 \, \Rightarrow x= [value] .\) \(f_y = 0 \Rightarrow [value] = 0 \, \Rightarrow y= [value] .\) These equations have solution \( x=y= 0,\) hence the stationary point is \( (0, 0) .\) Then we calculate \( f_{xx}, \, f_{yy}, \, f_{xy} \, \text{ and } \, D:\) \( f_{xx} = [value] \) \( f_{yy} = [value] \) \( f_{xy} = [value] \) \( D = f_{xx}f_{yy}- (f_{xy})^2 = [value].\) At the point \( (0,0)\) we have \( f_{xx} = [value], \, f_{yy} = [value], \, f_{xy} = [value]\) and \( D = [value] .\) We have \(D>0 \) and \(f_{xx} >0\) so the stationary point is a local minimum.

Exercise

In each of these figures, what are the features at \(A\) and \(B\)?

1. [Image: Plot of a 3 dimensional surface]
\(A\) is a:
\(B\) is a:
2. [Image: Plot of a 3 dimensional surface]
\(A\) is a:
\(B\) is a:
Feedback: 1. From feature \(A\), depending on the direction we move in we can either gain or loose height. Hence, \(A\) is a saddle point. From feature \(B\), no matter what direction we move in we always gain height. Hence, \(B\) is a local minimum. 2. From feature \(A\), no matter what direction we move in we always loose height. Hence, \(A\) is a local maximum. From feature \(B\), depending on the direction we move in we can either gain or loose height. Hence, \(B\) is a saddle point.

Exercise

Verify that \((0,2)\) is a stationary of the function \(f(x,y) = [value]\) and find the stationary value \(f(0,2)\).

First, find the partial derivatives:

\(f_x(x,y) = \)

\(f_y(x,y) = \)

Now find the values of these partial derivatives at \(x = 0\), \(y = 2\):

\(f_x(0,2) = \)

\(f_y(0,2) = \)

Laslty, evaluate \(f\) at this point:

\(f(0,2) = \)
Feedback: We are given the function \(f(x,y) = [value]\) and are asked to verify that the point \((0,2)\) is a stationary point, and find the value of \(f\) there. First, we find the partial derivatives: \[\begin{aligned} f_x(x,y) &= [value] \\ f_y(x,y) &= [value] \end{aligned}\] Now we evaluate them at \(x = 0\), \(y = 0\). \[\begin{aligned} f_x(0,2) &= [value] \\ f_y(0,2) &= [value] \end{aligned}\] Since both partial derivatives are zero, we conclude that \((0,2)\) is indeed a stationary point. Lastly, we find the value of \(f\) there: \[ f(0,2) = [value] \]

Exercise

Locate the stationary points of

\[f(x,y) = [value].\]

First find the partial derivatives of \(f\):

\(\displaystyle \frac{\partial f}{\partial x} = \)

\(\displaystyle \frac{\partial f}{\partial y} = \)

Now solve simultaenously the equations \(\displaystyle \frac{\partial f}{\partial x} = 0\) and \(\displaystyle \frac{\partial f}{\partial y} = 0\).

Note: write your answer as a set of points. For example, if you found the stationary points \((0,2)\) and \((1,3)\), your input should be {[0,2], [1,3]}.

Stationary points:
Feedback: We are asked to find the stationary points of \(f(x,y) = [value]\). We begin by finding the partial derivatives of \(f\): \[\begin{aligned} \frac{\partial f}{\partial x} &= [value] \\ \frac{\partial f}{\partial y} &= [value] \end{aligned}\] Next, we find the points where both partial derivatives are zero. Note that each of them only depends on one variable, so that we can solve each equation separately. \[\begin{aligned} \frac{\partial f}{\partial x} = [value] = 0 &\implies x = \pm [value] \\ \frac{\partial f}{\partial y} = [value] = 0 &\implies y = [value] \end{aligned}\] Therefore, there are two points where both partial derivatives are zero: \(([value],[value])\) and \(([value],[value])\). These are the stationary points.

Exercise

Determine the nature of the stationary points of \(f(x,y) = [value]\), which are \(([value],[value])\) and \(([value],[value])\).

First, write down the three second partial derivatives:

\(f_{xx} = \)

\(f_{yy} = \)

\(f_{xy} = \)

Now complete the table below, determining the nature of the stationary points:

Point | \(([value],[value])\) | \(([value],[value])\)
\(f_{xx}\) |  |
\(f_{yy}\) |  |
\(f_{xy}\) |  |
\(D\) |  |
Nature |  |
Feedback: We are asked to find the nature of the stationary points of \(f(x,y) = [value]\), which are given to be \(([value],[value])\) and \(([value],[value])\). First, we find the three second partial derivatives: \[\begin{aligned} f_{xx} &= [value] \\ f_{yy} &= [value] \\ f_{xy} &= [value] \end{aligned}\] Now we work through each stationary point separately. Recall that the formula for \(D\) is given by \[ D = f_{xx} \cdot f_{yy} - (f_{xy})^2. \] At \(([value],[value])\), we have: \[\begin{aligned} f_{xx}([value],[value]) &= [value] \\ f_{yy}([value],[value]) &= [value] \\ f_{xy}([value],[value]) &= [value] \end{aligned}\] It follows that \(D = [value] \cdot [value] - ([value])^2\) is [value]. Using the sign of \(D\) and \(f_{xx}\), we conclude that \(([value],[value])\) is a [value]. At \(([value],[value])\), we have: \[\begin{aligned} f_{xx}([value],[value]) &= [value] \\ f_{yy}([value],[value]) &= [value] \\ f_{xy}([value],[value]) &= [value] \end{aligned}\] It follows that \(D = [value] \cdot [value] - ([value])^2\) is [value]. Using the sign of \(D\) and \(f_{xx}\), we conclude that \(([value],[value])\) is a [value].

Example

Find a second stationary point of \(f(x,y)=8x^2+6y^2-2y^3+5\).

Solution

\(f_x=16x\) and \(f_y\equiv 6y(2-y)\).

From this we note that \(f_x=0\) when \(x=0\), and \(f_y=0\) when \(y=0\).

So \(x=0,\ y=0\)

i.e. \((0,0)\) is a second stationary point of the function.

It is important when solving the simultaneous equations \(f_x=0\) and \(f_y=0\) to find stationary points not to miss any solutions. A useful tip is to factorise the left-hand sides and consider systematically all the possibilities.

Example

Locate the stationary points of \(f(x,y)=x^4+y^4-36xy\)

Solution

First we write down the partial derivatives of \(f(x,y)\)

\(\dfrac{\partial f}{\partial x}=4x^3-36y=4(x^3-9y)\)

\(\dfrac{\partial f}{\partial y}=4y^3-36x=4(y^3-9x)\)

Now we solve the equations \(\dfrac{\partial f}{\partial x}=0\) and \(\dfrac{\partial f}{\partial y}=0\):

\(x^3-9y = 0 \\ \) (i)

\(y^3-9x = 0 \\ \) (ii)

From (ii) we obtain

\(x = \dfrac{y^3}{9}\) (iii)

Now substitute from (iii) into (i)

\(\dfrac{y^9}{9^3}-9y = 0\\ \) \(\Rightarrow y^9-9^4y = 0\\ \)

\(\Rightarrow y(y^8-3^8) = 0 \) (removing the common factor) \( \\ \)

\(\Rightarrow y(y^4-3^4)(y^4+3^4) = 0 \) (using the difference of two squares)\( \\ \)

We therefore obtain as the only solutions:

\(y=0\) or \(y^4-3^4 = 0\) (since \(y^4+3^4\) is never zero) \( \\ \)

The last equation implies:

\((y^2-9)(y^2+9) = 0\) (using the difference of two squares) \( \\ \)

therefore \(y^2=9\) and \(y =\pm\ 3\).

Now, using (iii): when \(y=0,\ x=0\), when \(y=3,\ x=3\), and when \(y=-3,\ x=-3\).

The stationary points are \((0,0), (-3,-3)\) and \((3,3)\).

Example

Locate the stationary points of \(f(x,y)=x^4+y^4-36xy\)

Solution

\(\dfrac{\partial f}{\partial x}=3x^2-3,\\ \dfrac{\partial f}{\partial y}=2y-6\)

Now solve simultaneously the equations \(\dfrac{\partial f}{\partial x}=0\) and \(\dfrac{\partial f}{\partial y}=0\):

\(3x^2-3=0\) and \(2y-6=0\).

Hence \(x^2=1\) and \(y=3\), giving stationary points at \((1,3)\) and \((-1,3)\).

Key Point

The Partial Derivative of \(\pmb f\) with respect to \(\pmb x\)

For a function of two variables \(z=f(x,y)\) the partial derivative of \(f\) with respect to \(x\) is denoted by \(\dfrac{\partial f}{\partial x}\) and is obtained by differentiating \(f(x,y)\) with respect to \(x\) in the usual way but treating the \(y\)-variable as if it were a constant.

Alternative notations for \(\dfrac{\partial f}{\partial x}\) are \(f_x(x,y)\) or \(f_x\) or \(\dfrac{\partial z}{\partial x}\)

Example

Find \(\dfrac{\partial f}{\partial x}\) for
1. \(f(x,y)=x^3+x+y^2+y\),
2. \(f(x,y)=x^2y+xy^3\).

Solution

1. \(\dfrac{\partial f}{\partial x}=3x^2+1+0+0=3x^2+1\)
2. \(\dfrac{\partial f}{\partial x}=2x\times y+1\times y^3=2xy+y^3\)

For most functions the procedures described above enable us to distinguish between the various types of stationary point. However, note the following example, in which these procedures fail.

Given\(f(x,y)=x^4+y^4+2x^2y^2.\) \(\dfrac{\partial f}{\partial x}=4x^3+4xy^2,\dfrac{\partial f}{\partial y}=4y^3+4x^2y,\) \(\dfrac{\partial^2f}{\partial x^2}=12x^2+4y^2,\dfrac{\partial^2f}{\partial y^2}=12y^2+4x^2, \dfrac{\partial^2f}{\partial x\partial y}=8xy\)

Location:

The stationary points are located where \(\dfrac{\partial f}{\partial x}=\dfrac{\partial f}{\partial y}=0,\) that is, where
\(4x^3+4xy^2=0\) and \(4y^3+4x^2y=0\). A simple factorisation implies \(4x(x^2+y^2)=0\) and \(4y(y^2+x^2)=0\). The only solution which satisfies both equations is \(x=y=0\) and therefore the only stationary point is \((0,0)\).

Nature:

Unfortunately, all the second partial derivatives are zero at \((0,0)\) and therefore \(D=0\), so the test, as described in the Key Point fails to give us the necessary information.

However, in this example it is easy to see that the stationary point is in fact a local minimum. This could be confirmed by using a computer generated graph of the surface near the point \((0,0)\). Alternatively, we observe \(x^4+y^4+2x^2y^2\equiv (x^2+y^2)^2\) so \(f(x,y)\ge 0\), the only point where \(f(x,y)=0\) being the stationary point. This is therefore a local (and global) minimum.

Exercise

Find the stationary points for the function \(f(x,y) = [value]\) .

For each stationary point classify it as either a local minimum, a local maximum or a saddle point.

1. \(\dfrac{\partial f}{\partial x} = \dfrac{\partial f}{\partial y} = 0\) at \((x, y)\) =