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Relative Minimums and
Maximums

In this section we are going to extend one of the more important ideas from Calculus I into
functions of two variables
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This in fact will be the topic of the following two sections as well
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Recall as well that we will often use the word extrema to refer to both minimums
and maximums
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So, for the sake of completeness here is the definition of relative minimums and
relative maximums for functions of two variables
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A function f ( x, y ) has a relative minimum at the point ( a,b ) if f ( x, y ) ≥ f ( a,b ) for
all points ( x, y ) in some region around ( a,b )
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A function f ( x, y ) has a relative maximum at the point ( a,b ) if f ( x, y ) ≤ f ( a,b ) for
all points ( x, y ) in some region around ( a,b )
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It only says that in some region around the point ( a,b ) the function will
always be larger than f ( a,b )
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Likewise, a relative maximum only says that around ( a,b ) the function will always
be smaller than f ( a,b )
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Next we need to extend the idea of critical points up to functions of two variables
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We have a similar definition for critical points of functions of two variables
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2
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To see the equivalence in the first part let’s start off with ∇f = 0 and put in the definition of each
part
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In
fact, we will use this definition of the critical point more than the gradient definition since it will
be easier to find the critical points if we start with the partial derivative definition
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If only one
of the first order partial derivatives are zero at the point then the point will NOT be a critical
point
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Fact
If the point ( a,b ) is a relative extrema of the function f ( x, y ) then ( a,b ) is also a critical
point of f ( x, y ) and in fact we’ll have ∇f ( a, b ) = 0
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Let’s start off by defining g ( x ) = f ( x,b ) and suppose that f ( x, y ) has a relative extrema at

( a,b )
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By Fermat’s Theorem we then know that g′ ( a ) = 0
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If we now define h ( y ) = f ( a, y ) and going through exactly the same process as above we will
see that f y ( a,b ) = 0
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Note that this does NOT say that all critical points are relative extrema
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To see this let’s consider the function

f ( x, y ) = xy

The two first order partial derivatives are,

f x ( x, y ) = y

f y ( x, y ) = x

The only point that will make both of these derivatives zero at the same time is ( 0, 0 ) and so

( 0, 0 )

is a critical point for the function
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Note that the axes are not in the standard orientation here so that we can see more clearly what is
happening at the origin, i
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at ( 0, 0 )
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However, if we start at the
origin and move into either of the quadrants where x and y have the opposite sign then the
function decreases
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Therefore, there is no way
that ( 0, 0 ) can be a relative extrema
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While we have to be careful to not misinterpret the results of this fact it is very useful in helping
us to identify relative extrema
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The fact tells us
that all relative extrema must be critical points so we know that if the function does have relative
extrema then they must be in the collection of all the critical points
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So, once we have all the critical points in hand all we will need to do is test these points to see if
they are relative extrema or not
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Fact
Suppose that ( a,b ) is a critical point of f ( x, y ) and that the second order partial derivatives are
continuous in some region that contains ( a,b )
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1
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2
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3
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4
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Other techniques would need to be used to classify the critical point
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Also note that

we aren’t going to be seeing any cases in this class where D = 0
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Let’s see a couple of examples
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Solution
We first need all the first order (to find the critical points) and second order (to classify the
critical points) partial derivatives so let’s get those
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Critical points will be solutions to the system of equations,

f x = 3x 2 − 3 y = 0
f y = 3 y − 3x = 0
2

This is a non-linear system of equations and these can, on occasion, be difficult to solve
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We can solve the first equation for y as follows,

3x 2 − 3 y = 0

⇒

y = x2

Plugging this into the second equation gives,

3( x

)

2 2

− 3 x = 3x ( x3 − 1) = 0

From this we can see that we must have x = 0 or x = 1
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x = 0 : y = 02 = 0

⇒

x = 1: y = 12 = 1

⇒

( 0, 0 )
(1,1)

So, we get two critical points
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To do this we will need
D
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D ( x, y ) = f x x ( x, y ) f y y ( x, y ) − ⎡⎣ f x y ( x, y ) ⎤⎦
= ( 6 x )( 6 y ) − ( −3)

2

2

= 36 xy − 9
To classify the critical points all that we need to do is plug in the critical points and use the fact
above to classify them
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(1,1) :
D = D (1,1) = 36 − 9 = 27 > 0

f x x (1,1) = 6 > 0

For (1,1) D is positive and f x x is positive and so we must have a relative minimum
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Notice that in order to get a better visual we used a somewhat nonstandard orientation
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Example 2 Find and classify all the critical points for f ( x, y ) = 3x 2 y + y 3 − 3x 2 − 3y 2 + 2
Solution
As with the first example we will first need to get all the first and second order derivatives
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The equations that we’ll need to solve this time are,

6 xy − 6 x = 0

3x 2 + 3 y 2 − 6 y = 0
These equations are a little trickier to solve than the first set, but once you see what to do they
really aren’t terribly bad
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Be careful to not just cancel
the x from both sides
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To find the critical points we can plug these (individually) into the second equation and solve for
the remaining variable
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To do this we’ll need the general formula for
D
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( 0, 0 )
( 0, 2 )
(1,1)
( −1,1)

: Relative Maximum
: Relative Minimum
: Saddle Point
: Saddle Point

Here is a graph of the surface for the sake of completeness
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Example 3 Determine the point on the plane 4 x − 2 y + z = 1 that is closest to the point
( −2, −1,5)
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In order to do this example we
are going to need to first come up with the equation that we are going to have to work with
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The distance between this point and
the point in question, ( −2, −1,5 ) , is given by the formula,

d=

( x + 2 ) + ( y +1) + ( z − 5)
2

2

2

What are then asking is to find the minimum value of this equation
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There are a couple of issues with this equation
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This is easy to fix however
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So, in order to make our life a little easier let’s notice that
2
finding the minimum value of d will be equivalent to finding the minimum value of d
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We are being asked to find the closest point on the plane
to ( −2, −1,5 ) and that is not really the same thing as what we’ve been doing in this section
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Hopefully, it does make sense from a physical standpoint that there will be a closest
point on the plane to ( −2, −1,5 )
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So, let’s go through the process from the first and second example and see what we get as far as
relative minimums go
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We’ll need the derivatives first
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D = 34 (10 ) − ( −16 ) = 84 > 0
2

So, in this case D will always be positive and also notice that f x x = 34 > 0 is always positive and
so any critical points that we get will be guaranteed to be relative minimums
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This will mean solving the system
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1
1
x = (16 y − 36 ) = ( 8 y − 18 )
17
34
Now, plug this into the second equation and solve for y
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25
,
−
So, it looks like we get a single critical point : ( − 34
21
21 )
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We can find the z coordinate by plugging into the
equation of the plane as follows,

⎛ 34 ⎞ ⎛ 25 ⎞ 107
z = 1− 4⎜ − ⎟ + 2⎜ − ⎟ =
⎝ 21 ⎠ ⎝ 21 ⎠ 21
25 107
So, the point on the plane that is closest to ( −2, −1,5 ) is ( − 34
,
−
21
21 , 21 )

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