Chapter 10 Multi-Variable Functions
10.1 Functions of Several Independent Variables
In the previous chapters we worked with functions containing one independent variable. Now we will
expand this concept and include more than one variable in a function. Let's look at this in the form of an
example.
Scoot-It is a motorized scooter company that produces one type of scooter called the explorer. If
each explorer scooter sells for $7500 then the company's weekly revenue function, in hundreds of dollars,
can be modeled by R( x)  75 x where x represents the number of explorer scooters sold each week. This
revenue function is a function of one independent variable. Suppose, however, the company decides to
expand its line of scooters and begins to produce the discovery scooter, a model the company sells for
$5000. If we let y be the number of discovery scooters sold each week then the company's weekly
revenue function, in hundreds of dollars, for selling both types of scooters is R( x, y )  75 x  50 y . This
second revenue function is a function of two independent variables, namely x and y, and R(x,y) is called
the dependent variable. We can use this revenue function to determine the amount of revenue generated
when 4 explorer scooters and 6 discovery scooters are sold by evaluating R(4,6) as shown below.
R(4,6)  75(4)  50(6)  600
Thus, Scoot-It will make $60,000 in revenue when 4 explorer scooters and 6 discovery scooters are sold.
Example 10-____: A local music hall is selling ticket to Friday’s concert. The matinee show costs $15
per person and the evening show costs $25 per person. If x is the number of matinee tickets purchased
and y is the number of evening tickets purchased for the Friday concert, find the revenue function R(x,y)
for Friday's concert then evaluate and interpret R(125, 200).
Solution: Since revenue is the selling price times the number of tickets sold, the revenue function for
Friday's concert is R( x, y )  15 x  25 y and R(125, 200)  15(125)  25(200)  6875 . This means that
when 125 tickets are sold for the matinee show and 200 tickets are sold for the evening show, the music
hall generates a revenue of $6,875.END OF SOLUTION
Example 10-____: The Scoot-It motorized scooter company has determined that its demand equation for
the two types of scooters are given by
p  35  3x  4 y (p  price of explorer scooter in hundres of dollars)
q  20  2 x  y (q  price of discovery scooter in hundres of dollars)
where x is the number of explorer scooters sold per week and y is the number of discovery scooters sold
per week. Find the weekly revenue function R(x,y) and then evaluate and interpret R(10, 18).
Solution: Since revenue is the price times the number of items sold the revenue function for Scoot-It is
R ( x, y )  p  x  q  y
 (35  3x  4 y ) x  (20  2 x  y ) y
 35 x  3x 2  4 xy  20 y  2 xy  y 2
 y 2  3x 2  2 xy  20 y  35 x
So
R(10,18)  (18)2  3(10)2  2(10)(18)  20(18)  35(10)
 1094
This means that when 10 explorer scooters and 18 discovery scooters are produced and sold Scoot-It
earns $109,400 in revenue. END OF SOLUTION
Example 10-____: If the Scoot-It motorized scooter company's weekly cost function, in hundreds of
dollars, is given by C ( x, y )  8.5 x  6 y  400 and the revenue function is the same from example 10___
above, determine the company's profit function P(x,y) then evaluate and interpret P(10,18).
Solution: Since profit = revenue - cost we can obtain the following profit function for Scoot-It.
P( x, y)  ( y 2  3x 2  2 xy  20 y  35 x)  (8.5x  6 y  400)
 y 2  3x 2  2 xy  20 y  35 x  8.5x  6 y  400
 y 2  3x 2  2 xy  14 y  26.5x  400
So
P(10,18)  (18)2  3(10)2  2(10)(18)  14(18)  26.5(10)  400
 501
Thus, when 10 explorer scooters and 18 discovery scooters are produced and sold, Scoot-It earns $50,100
in profit.END OF SOLUTION
10.2 Cobb–Douglas Production Function
Production functions help companies estimate how many goods can be produced with a given
number of inputs. A commonly used production function is one of the form Q  f ( L, K )  aLb K c and is
called the Cobb–Douglas production function, named after the two founders. In the Cobb-Douglas
production function, L is the number of units of labor, K is the number of unit of capital, and a, b, c are
constants.
Example 10-____: Suppose a metal manufacturing company has a Cobb-Douglas production function,
in hundreds of pounds, of f ( L, K )  10L0.25 K 0.75 where L is the number of hours of labor and K is the
dollar amount of capital invested. If the company uses 2,000 hours of labor and $1500 in capital, how
many pounds of metal will be produced?
Solution: To find the number of pounds of metal produced we need to evaluate
f (2000, 750)  10(2000)0.25 (1500)0.75
 10(6.6874)(241.0285)
 16119
Thus, the manufacturing company will produce approximately 1,611,900 pounds of metal when 2000
hours of labor and $1500 in capital is invested. End of Solution
The advantages of using the Cobb-Douglas production function for estimating production are many and
are widely used in empirical work. One advantage of the Cobb-Douglas function is that returns to scale
can be determined by simply summing the exponents on the two inputs, labor and capital. The chart
below shows how to determine the returns to scale.
Properties of the Cobb-Douglas Production Function
Given a Cobb-Douglas production function f ( L, K )  aLb K c if
 b + c = 1 then the function displays constant return to scale. That is,
increasing labor and capital by a factor of p causes production to increase by
the same factor p.
 b + c > 1 then the function displays increasing returns to scale. That is an
increase in labor and capital by a factor p causes production to increase by
more than the factor p.
 b + c < 1 then the function displays decreasing returns to scale. That is an
increase in labor and capital by a given factor p causes production to increase
by less than the factor p.
Let’s look at this in the form of an example. If a company’s Cobb-Douglas production function is
represented by f ( L, K )  L0.3 K 0.7 then f (2,4)  20.340.7  3.249 . If we increase labor and capital by a
factor of 3 then we would expect production to increase by a factor of 3 because 0.3 + 0.7 = 1.
f (6,12)  60.3120.7  9.747 .
We notice that 3(3.249) = 9.747. Thus, we have confirmed the production function displays constant
returns to scale.
Now suppose a company has a Cobb-Douglas production function of f ( L, K )  L0.4 K 0.8 then
f (2,4)  20.4 40.8  4 . Once again, if we increase labor and capital by a factor of 3 then we would expect
production to increase by more than a factor of 3 because 0.4 + 0.8 = 1.2.
f (6,12)  60.4120.8  14.949
Since 14.949 > (4)(3) = 12 the production function displays increasing returns to scale.
Lastly, suppose a company has a Cobb-Douglas production function of f ( L, K )  L0.2 K 0.5 , then
f (2,4)  20.2 40.5  2.297 . If labor and capital are increased by a factor of 3, then we would expect
production to increase by less than a factor of 3 because 0.2 + 0.5 = 0.7 < 1.
f (6,12)  60.2120.5  4.957
Since 4.957 < (2.297)(3) = 6.891 the production function displays decreasing returns to scale.
Example 10-______: A sporting goods manufacturing company has a Cobb-Douglas production
function of f ( L, K )  2L.36 K .43 , where f ( L, K ) is the number of units produced. If the company invests
800 hours of labor and $3000 in capital, how many sporting good units will be produced? What type of
returns to scale does the production function display?
Solution: Evaluating f (800,3000) we get
f (800,3000)  2(800)0.36 (3000)0.43  694 .
Thus, the sporting goods manufacturing company can expect to produce 694 units of sporting goods when
800 hours of labor and $3000 in capital are invested. Since the sum of the exponents on the inputs are
less than 1, that is 0.36 + 0.43 = 0.79, the production function displays a decreasing returns to scale.
End of solution.
Returns to scale are important for determining how many firms will populate an industry. When
increasing returns to scale exist, one large firm will produce more cheaply than two small firms. Small
firms will thus have a tendency to merge to increase profits, and those that do no merge will eventually
fail. On the other hand, if an industry has decreasing returns to scale, a merger of two small firms to
create a large firm will cut output, raise average costs, and lower profits. In such industries, many small
firms should exist rather than a few large firms.
(http://ingrimayne.saintjoe.edu/econ/TheFirm/ProductionFunct.html)
10.3 The 3-Dimensional Coordinate Plane
Now that we are familiar with functions of two variables it is time to look at the graphs created by
these functions. Since we have two independent variables and one dependent variable, the 3-dimensional
coordinate plane consists of three axes, all of which are perpendicular to one another. The points in a 3dimensional coordinate system are represented with an ordered triple (x, y, z). The point (1, 2, 3) is
graphed below.
Figure 10-_____ A plot of the point (1, 2, 3).
The graph of a function f(x,y) is called a surface and are typically very difficult to graph by hand. Thus,
in these modules we will use a 3-dimensional graphing applet to graph any function of two variables.
Some examples of surfaces are shown in figure 10-___ below.
Figure 10-_____ Graphs of various functions of two variables.
Example 10-___: Find the domains of the following functions
f ( x, y ) 
f ( x, y )  x  y
a)
f ( x, y )  ln(2 x  y )
b)
x y
x y
c)
Solution:
a) For the function
f ( x, y )  x  y
f ( x)  x  y
is all x such that
the inequality x  y  0 must hold true. Therefore the domain of
x  y .
f ( x, y ) 
b) Since the denominator of a rational function can not be zero, the domain of
x  y  0 or x   y .
such that
c) Since the arguement of a logarithmic function must always be greater than zero, that is
domain of
f ( x, y )  ln(2 x  y )
is all x such that
2x  y .
x y
x y
is all x
2 x  y  0 , the
10.4 First Order Partial Derivatives
In section 10-1 we were given the cost, revenue, and profit functions for the Scoot-It motorized scooter
company. In addition, we found P(10,18) =501 and interpreted this to mean that when 10 explorer
scooters and 18 discovery scooters are sold the company earns $50,100 in profit. Now supose the
company's owners need to determine if the production of more scooters will result in more profit, and if
more profit is obtained, which type of scooter would produce more profit. To do this type of analysis, we
must take first order partial derivatives. That is, we need to fix the production of one type of scooter
and determine the amount of addition profit obtained when the production of the other scooter increases
by one. So if we take the profit function
P(x, y)  y2  3x2  2xy 14 y  26.5x  400
and fix the number of discover scooters produced at 18 but let the production of explorer scooters vary the
following profit function is formed.
P( x,18)  (18) 2  3x 2  2 x(18)  14(18)  26.5 x  400
 3x 2  62.5 x  176
Since P(x) contains a single variable, we can find the marginal profit function by taking the derivative of
P(x,18)  3x2  62.5x 176 with respect to x as shown below.
Px ( x,18)  6x  62.5
Substituting x = 10 into the marginal profit function will approximate the amount of profit that is obtained
when one additional explorer scooter is produced.
Px (10,18)  6(10)  62.5  2.5
Thus, at a production level of 10 explorer and 18 discovery scooters, Scoot-It's profit will increase by
approximately $250 when one additional explorer scooter is produced. We can do the same analysis if we
fix the number of explorer scooters produced to 10 and allow the number of discovery scooters to vary.
This gives us a profit function in terms of the single variable y.
P(10, y)  y 2  3(10)2  2(10) y  14 y  26.5(10)  400
 y 2  34 y  435
Now we can find the marginal profit function for the discovery scooters.
Py (10, y )  2 y  34
Substituting y = 18 into the marginal profit function will approximate the amount of profit that is obtained
when one additional discover scooter is produced.
Py (10,18)  2(18)  34
 70
Thus, at a production level of 10 explorer and 18 discovery scooters, Scoot-It's profit will increase by
approximately $7,000 when the 19th discovery scooter is produced. From our analysis, we conclude that
Scoot-It's marginal profit is higher for the discovery soooters when the production level is 10 explorer and
18 discovery scooters. It is important to notice that when the partial derivative of P(x,y) was taken with
respect to x it was denoted by
y it was denoted by
Py ( x, y )
Px ( x, y) and when the partial derivative of P(x,y) was taken with respect to
.
Example 10-___: Given f ( x, y)  3x 2 y  2 y 3 x  100 find
f (4,7)
f x (4,7) and f y (4, 7) .
Solution: When finding x
we need to substitute y = 7 into f(x,y), take a partial derivative with
respect to x and then substitute x = 4 into the resulting function.
f ( x, 7)  3 x 2 (7)  2(7)3 x  100  21x 2  686 x  100
f x ( x, 7)  42 x  686
f x (4, 7)  42(4)  686  518
f (4, 7)
Now to find y
we need to substitute x = 4 into f(x,y), take a partial derivative with respect to y and
then substitute y = 7 into the resulting function.
f (4, y )  3(4) 2 y  2 y 3 (4)  100  48 y  8 y 3  100
f y (4, y )  48  24 y 2
f y (4, 7)  48  24(7) 2  1128
END OF SOLUTION
Let’s look at the previous example geometrically. When y is fixed at 7, that is when you substitute y = 7
into the function, the surface f ( x, y)  3x 2 y  2 y 3 x  100 is cut parallel to the x-axis 7 units from the
origin. The cross-section that is formed by this cut is a curve as shown in the figure below.
Figure 10-_____
Use the applet to draw f ( x, y)  3x 2 y  2 y 3 x  100 and the plane y = 7
When we take a partial derivative with respect to x we are finding the slope of the lines tangent to this
cross-section. If we were to draw a tangent line at x = 4, we know from our calculations above the
tangent line would have a slope of –518. A different curve is formed when f ( x, y)  3x 2 y  2 y 3 x  100
is cut by the plane x = 4. The intersection of x = 4 and the surface, along with the resulting cross-section
is shown in the figure below.
Figure 10-_____
Use the applet to draw f ( x, y)  3x 2 y  2 y 3 x  100 and the plane x = 4
If we were to draw a tangent line on these curve where y = 7 we know that the slope of that tangent line
would be –1128.
We are now ready to find partial derivatives of f ( x, y ) without reducing it to a function of one variable.
Example 10-___: Find
f x ( x, y) and f y ( x, y ) for each of the following functions
2
2
3
a) f ( x, y)  3x y  6x  2 y
2y 3
2 2
b) f ( x, y)  e y  ln(x y )
3
x y 2
c) f ( x, y)  (3x y  e )
2
Solution:
a) To find
need to find
f x ( x, y) we need to hold y constant and take the derivative of f(x,y) with respect to x, thus we


 
 
d
d
d
3x 2 y 
6 x2 
2 y3
dx
dx
dx
Since the derivative of a constant times a function is the constant times the derivative of the function we
obtain the following derivatives for each of the terms.
d
d
d
f x ( x, y ) 
3x 2 y 
6x2 
2 y3
dx
dx
dx
d 2
d
 d

 3 y  x2   6
x  2 y 3 (1)
dx
dx
 dx 


 


 
 3 y (2 x)  6(2 x)  2 y 3 (0)
 6 xy  12 x
f ( x, y)
To find y
we need to treat x as a constant and take the derivative of f(x,y) with respect to y, thus
we need to find
d
d
d
3x 2 y    6 x 2    2 y 3 

dy
dy
dy
Taking each of these derivatives we get the following.
f y ( x, y ) 


 

d
d
d
3x 2 y 
6x2 
2 y3
dy
dy
dy

 d 
d
d
 3x 2  y   6 x 2 1  2 ( y 3 )
dy
dx
 dy 
 3x 2 (1)  6 x 2 (0)  2(3 y 2 )
 3x 2  6 y 2
b) To find f x ( x, y ) we need to treat y as a constant and take the derivative of each term with respect to x.
To do this we will need to apply the derivative rule for logarithmic functions.
f x ( x, y ) 





Now let's find
f y ( x, y)




d 2y 3
d
e y 
ln( x 2 y 2 )
dx
dx
d
d
y 3e 2 y 1 
ln( x 2 y 2 )
dx
dx
d 2 2
x y
y 3e 2 y (0)  dx 2 2
x y
d 2
y2
x
dx
x2 y 2
2x
x2
2
x




 
. To take this partial derivative we will need to apply the product rule to find
d 2y 3
e y 
dx
as well as use derivative rules for exponential and logarithmic functions.
.
d 2y 3
d
e y    ln( x 2 y 2 ) 

dy
dy
d
d
d
 y 3  e 2 y   e 2 y  y 3    ln( x 2 y 2 ) 
dy
dy
dy
f y ( x, y ) 
 y 3  2e 2 y   e 2 y (3 y 2 ) 
 2 y 3e 2 y  3 y 2 e 2 y 
2x2 y
x2 y 2
2
y
c) To find f x ( x, y ) we will need to use the chain rule and as well as the derivative rules for exponential
functions.


2
2
d
3x 3 y  e x y
dx
2
d 2 
 d
 2 3x3 y  e x y  3x3 y  e x y 
dx
 dx

f x ( x, y ) 



  3 y dxd x  dxd e 
 2  3x y  e  3 y (3x )  e (2 xy ) 
 2  3x y  e  9 x y  2 xye 
 2 3x3 y  e x
Now let's find
2
y
3
x2 y
3
x2 y
x2 y
3
2
x2 y
x2 y
2
f y ( x, y)
f y ( x, y ) 

2
d
3x3 y  e x y
dy

 2 3x3 y  e x
2
y

2
  dyd 3x y  dyd e
3
x2 y




  3x dyd y  dyd e 
 2  3x y  e  3x (1)  e ( x ) 
 2  3x y  e  3x  x e 
 2 3x3 y  e x
2
x2 y
y
3
3
x2 y
2
x2 y
3
x2 y
2
2 x2 y
2
10.4 Competitive vs. Complementary Commodities
Two commodities are said to be competitive, or substitute, if a decrease in the demand of one
commodity produces an increase in the demand of the other commodity. Suppose two brands of coffee,
Jitters and Shakers, are competitive commodities. Let f ( p, q) represent the number of 20oz bags of Jitters
brand coffee demanded and g ( p, q) represent the number of 20oz bags of Shakers brand coffee
demanded where p is the price per bag of Jitters brand coffee and q is the price per bag of Shakers brand
coffee. If the price of Jitters brand coffee increases while the price of Shakers brand coffee is held
constant then the demand for Jitters brand coffee will typically decrease, that is f p ( p, q)  0 , and since the
commodities are competitive, the demand for Shakers brand coffee will increase, that is g p ( p, q)  0 .
Likewise, if the price of Shakers brand coffee increases while the price for Jitters brand coffee is held
constant, then the demand for Shakers brand coffee will typically decrease, that is g q ( p, q)  0 , and since
the commodities are competitive, the demand for Jitters brand coffee will increase, that is f q ( p, q)  0 .
Two commodities are said to be complementary if a decrease in the demand of one commodity
produces a decrease in the demand of the other commodity. Suppose digital video disk (DVD) players
and digital video disks (DVDs) are complementary commodities. Let f ( p, q) represent then number of
DVD players demanded and g ( p, q) represent the number of DVDs demanded where p is the price of
each DVD player and q is the price of each DVD. If the price of a DVD player increases and the price of
a DVD is held constant, then the demand for DVD players will decrease, f p ( p, q)  0 and since these two
commodities are complementary, the demand for DVDs will also decrease, g p ( p, q)  0 . Similarly, if the
price for DVDs increases and the price for DVD players is held constant, then the demand for DVDs will
decrease, g q ( p, q)  0 , and since the commodities are complementary, the demand for DVD players will
decrease, f q ( p, q)  0 . This leads us to the following test for complementary and competitive
commodities.
Test for Competitive and Complementary Commodities
Let f ( p, q) represent then number of commodity A demanded and g ( p, q) represent the
number of commodity B demanded where p is the unit price commodity A and q is the unit
price of commodity B.
Partial Derivatives
Commodities A and B
f q ( p, q)  0 and g p ( p, q)  0
Competitive
f q ( p, q)  0 and g p ( p, q)  0
Complementary
f q ( p, q)  0 and g p ( p, q)  0
Neither
f q ( p, q)  0 and g p ( p, q)  0
Neither
Example 10-_____: If f ( p, q)  150  6 p  5q is the demand function for commodity A and
g ( p, q)  250  3 p  5q is the demand function for commodity B, determine if the two commodities are
complementary, competitive, or neither.
Solution: We need to find f q ( p, q) and g p ( p, q) .
f q ( p, q)  5 and g p ( p, q)  6
Thus, these commodities are neither competitive or complementary.
10.5 Second Order Partial Derivatives
A second order partial derivative is the derivative of the first order partial derivative and they are typically
denoted in the following ways.
f xx ( x, y )  The first and second partial derivatives are taken with respect to x.
f yy ( x, y )  The first and second partial derivatives are taken with respect to y.
f xy ( x, y )  The first partial derivative is taken with respect to x and the second partial
derivative is taken with respect to y.
f yx ( x, y )  The first partial derivative is taken with respect to y and the second partial
derivative is taken with respect to x.
2
3
Example 10-___: Given f ( x, y)  3x y  2 y x  40 find all four second order partial derivatives.
f ( x, y)
Solution: To find the second order partial derivatives we first need to find f x ( x, y ) and y
.
f y ( x, y)  3x 2  6 y 2 x
f x ( x, y )  6 xy  2 y 3
f ( x, y )  6 xy  2 y 3
To find f xx ( x, y) we will take the partial derivative of x
with respect to x.
f xx (x, y)  6 y
To find
f xy ( x, y)
we will take the partial derivative of
f x ( x, y )  6 xy  2 y 3
f x ( x, y )  6 x  6 y 2
with respect to y.
To find
f yy ( x, y)
f y ( x, y)  3x 2  6 y 2 x
we will take the partial derivative of
with respect to y.
f yy ( x, y )  12 yx
To find
f yx ( x, y)
f y ( x, y)  3x 2  6 y 2 x
we will take the partial derivative of
f yx ( x, y)  6 x  6 y
with respect to x.
2
End of Solution
f xy ( x, y)  f yx ( x, y)
In the example above we notice that
. Although this does not hold true for all second
order partial derivatives it is very difficult to find an application where this equation is not true. Thus, for
f ( x, y)  f yx ( x, y)
all of the functions in these modules the equation xy
will hold true.
10.6 Relative Extrema and Saddle Points
Just as we found relative maximum and minimum for functions of one variable, we will now
investigate how relative maximum and minimum points are found for functions of two variables.
First, however, we will make the assumption that all the functions we encounter will have second order
partial derivatives that exist in some circular region in the xy-plane. This simply ensures that we are
dealing with a smooth surface.
The first step in finding relative extrema for functions of two variables is to find the critical point(s).
Critical point(s) are found using the definition below.
Definition of Critical Point
The point (a,b) is called a critical point of f(x,y) if f x ( x, y )  0 and f y ( x, y )  0 .
Example 10-______: Determine the critical point for f ( x, y)  2x2  3 y 2  4x  3 y  5
Solution: First we need to find f x ( x, y ) and f y ( x, y )
f y ( x, y)  6 y  3
f x ( x, y )  4 x  4
Now we find where f x ( x, y )  0 and f y ( x, y )  0 .
0  4x  4
1 x
0  6y  3
1
y
2
So the critical point for f ( x, y)  2x2  3 y 2  4x  3 y  5 is (1,  1 2) .

Example 10-______: Determine the critical point for f ( x, y)  x2  y 2  4xy  8x  10 y  5
Solution: First we need to find f x ( x, y ) and f y ( x, y ) then set them equal to zero.
f x ( x, y )  2 x  4 y  8
f y ( x, y )  2 y  4 x  10
0  2x  4 y  8
0  2 y  4 x  10
After we set these two first partial derivatives equal to zero we will need to solve for x and y using the
substitution method. To do this we will choose one of the first partial derivatives and solve it for either x
or y. Let’s choose the partial derivative with respect to x f x ( x, y )  0 and solve it for x.
0  2x  4 y  8
4 y  8  2x
2y  4  x
Now we will take this expression for x and substitute it into f y ( x, y )  0
0  2 y  4(2 y  4)  10
0  2 y  8 y  16  10
0  6 y  6
6  6 y
1 y
Now substitute Now substitute y=1 into 2y-4=x and we get
x  2(1)  4
x 24
x  2
2
2
Thus, the critical point for f ( x, y)  x  y  4xy  8x  10 y  5 is (2,1) .
1
1
Example 10-_______: Determine the critical point for f ( x, y)  x3  x 2  y 2  4 y  50 .
3
2
Solution: We need to find both first order partial derivatives, set them equal to zero, and then solve for x
and y.
f x ( x, y )  x 2  2 x
f y ( x, y )  y  4
0  x2  2 x
and
0 y4
0  x( x  2)
y4
x  0, 2
1
1
Thus the critical points for f ( x, y)  x3  x 2  y 2  4 y  50 are (0, 4) and (–2, 4). END
3
2
Now that we know how to find critical points for functions of two variables we are now ready to find
relative extrema for functions of two variables. When we found relative extrema for functions of one
variable we found where the first derivative was equal to zero and then performed a first derivative test to
locate relative extrema. The process for finding relative extrema for functions of two variables is very
similar. We will find the critical points and then perform a second derivative test to locate any relative
extrema. We will use the following second derivative test when finding relative extrema of functions of
two variables.
Second-Derivative Test for Relative Extrema
Given a function f ( x, y ) whose second-order partial derivatives exist in some
circular region containing the critical point (a,b) as its center and if
2
D  f xx (a, b)  f yy (a, b)   f xy (a, b)  then




If
If
If
If
D  0 and f xx (a, b)  0 , then f (a, b) is a relative maximum.
D  0 and f xx ( a, b)  0 , then f (a, b) is a relative minimum.
D  0 , the f(x,y) has a saddle point at (a,b).
D  0 then the test fails.
Example 10-______: Find all relative extrema and/or saddle points for
f ( x, y)  2x2  3 y 2  4x  3 y  5 .
Solution: Figure 10-____ graphs the given function. From this graph it is obvious that a relative
maximum exists at the critical point.
Figure 10-____:
Snag a graph of f ( x, y)  2x2  3 y 2  4x  3 y  5
First we need to find the critical point(s) for f ( x, y)  2x2  3 y 2  4x  3 y  5 . That is we need to find
where f x ( x, y )  0 and f y ( x, y )  0 .
f x ( x, y )  4 x  4
f y ( x, y )  6 y  3
0  4 x  4 and
x  1
0  6 y  3
1
y
2
 1
Thus, the critical point for the given function is at  1,  .
2

To determine if any relative extrema or saddle points exist we need to find the second partial derivatives.
f xx ( x, y)  4, f yy ( x, y)  6, f xy ( x, y)  f yx ( x, y)  0
Now we need to evaluate
D  f xx (a, b)  f yy (a, b)   f xy (a, b) 
 (4)  (6)  0
2
.
 24
Since D > 0 and f xx (a, b)  4  0 we conclude that a relative maximum exists at
 1, 1 2 , f (1, 1 2)    1, 1 2 ,7.75 .
End of Solution
Example 10-______: Find all relative extrema for f ( x, y)  x3  6 x 2 
1 2
y  4y  4 .
2
Solution: First we need to find all critical points for f(x,y).
f y ( x, y )  y  4
f x ( x, y )  3x 2  12 x
0  3x( x  4) and
0 y4
x  0, 4
y  4
Thus critical points exist at (0, –4) and (–4, –4). Now to find whether these critical points are relative
maximums, minimums, saddle points, or neither we need to find the second order partial derivatives.
f xx ( x, y)  6 x  12, f yy ( x, y)  1, f xy ( x, y)  f yx ( x, y)  0
Let’s first investigate (–4, –4) for relative extrema and saddle points. First we evaluate
2
D  f xx (a, b)  f yy (a, b)   f xy (a, b)  where a = –4, b = –4.
D  6(4)  12 1  02  12
Since D < 0, we conclude a saddle point exists at  4, 4, f (4, 4)   (4, 4,28) .
Now we determine if any relative extrema or saddle points exist at (0, –4).
D  [6(0)  12] 1  0  12 .
Since D > 0 and f xx (0, 4)  12  0 we determine that a relative minimum exists at
(0, 4, f (0, 4))  (0, 4, 4) . We can confirm our answer by graphing the given function as shown below
in figure 10-_____.
Figure 10-______: Need a shot snagged of the given function.
Example 10-____: An on-line fitness company sells two types of shoes, aerobic and running. The store
pays $30 for a pair of aerobic shoes and $45 for a pair of running shoes. The daily demand equations for
each type of shoe are as follows:
x  850  36 p  15q demand equation for aerobic shoes
y  1075  20 p  25q demand equation for running shoes
where p is the selling price for each pair of aerobic shoes and q is the selling price for each pair of running
shoes. What price should the store sell each model of shoe if they want to maximize their profits?
Solution: Since profit = revenue – cost we need to formulate a cost and revenue equation. Since we
want to find the selling price for each type of shoe we need to formulate functions in terms of p and q.
The revenue function is
R ( p, q )  x  p  y  q
 (850  36 p  15q ) p  (1075  20 p  25q )q
 850 p  36 p 2  15 pq  1075q  20 pq  25q 2
 30 p 2  25q 2  35 pq  850 p  1075q
The cost function in terms of x and y is C ( x, y)  30 x  45 y where x is the number of aerobics shoes sold
and y is the number of running shoes sold. To get the cost function in terms of p and q we need to use the
demand equations as shown below.
C ( p, q)  30(850  36 p  15q)  45(1075  20 p  25q)
 25500  1080 p  450q  48375  900 p  1125q
 180 p  675q  73875
Now that we have a cost and revenue function we can find the profit function as follows
P ( p, q )  R ( p, q )  C ( p, q )
 (30 p 2  25q 2  35 pq  850 p  1075q)  (180 p  675q  73875)
 30 p 2  25q 2  35 pq  850 p  1075q  180 p  675q  73875
 30 p 2  25q 2  35 pq  670 p  1750q  73875
Now to find the selling prices that will maximize profit we need to find the critical point and run the
second derivative test. First we will find the critical point.
Pp ( p, q )  60 p  35q  670
Pq ( p, q )  50q  35 p  1750
and
0  60 p  35q  670
0  50q  35 p  1750
To solve this we need to use the substitution method. We start by solving Pq ( p, q)  0 for q.
35 p  1750  50q
0.7 p  35  q
Now we substitute this expression for q into Pp ( p, q)  0 .
0  60 p  35(0.7 p  35)  670
0  60 p  24.5 p  1125  670
1795  35.5 p
50.56  p
So q  0.7(50.56)  35  70.39 . Now that we have the critical point of (50.56, 70.39) we need to make
sure that a relative maximum exists at this critical point. Thus we need to find the second order partial
derivatives.
Ppp ( p, q)  60, Pqq ( p, q)  50, Ppq ( p, q)  Pqp ( p, q)  35
2
Now applying the second derivative test by evaluating D  Ppp ( p, q)  Pqq ( p, q)   Ppq ( p, q)  we get the
following
D  (60)(50)  (35) 2
 3000  1225
 1775
Since D > 0 and Ppp ( p, q)  0 the second derivative test states that a relative maximum exists at the
critical point. Thus, to maximize profit, the shoe company should sell a pair of aerobic shoes for
$50.56 and a pair of running shoes for $70.39. End of Solution
Sample Quiz
1. A computer company sells two types of laser printers. The demand equations for each type of printer
are as follows:
p  0.5x  0.1y  179
demand for type A laser printer
q  0.09 x  0.08 y  238 demand for type B laser printer
where x is the number of type A laser printers sold and y is the number of type B laser printers sold. If it
costs the company $75 for each type A laser printer and $95 for each type B laser printer, find the
company’s profit function, P( x, y ) , then evaluate and interpret P(200, 250)
2. The productivity of a third world country is approximated by the function f ( L, K )  10L0.77 K 0.26 where
L is the units of labor and K is the units of capital invested in production. Evaluate and interpret
f (500,150) then determine the returns to scale for this country.
3. Find the domain of f ( x, y)  ln(2 x  y)
4. A music production company has released two new compact discs (CDs). The company’s revenue
function, in dollars, is given by
R( x, y)  1.5x2  y 2  4x  7 y  2xy  370
Evaluate and interpret Rx (125,100) .
5. Given f ( x, y)  (e2 y  x  ln x)3 find f x ( x, y ) and f y ( x, y ) .
6. An electronics company has two products whose demand functions are as follows:
f ( p, q)  150  2 p3  2q 2 demand for product A
g ( p, q)  175  3 p 2  5q3 demand for product B
Determine whether these two products are competitive, complementary, or neither.
7. Given f ( x, y)  xe xy find f xy ( x, y ) .
8. Find the critical point(s) for f ( x, y)  x3  y3  6xy .
9. Find all relative extrema and/or saddle point(s) for f ( x, y)  x3  y3  3xy  10 .
10. A home and garden shop sells two types of tomato plants. The shop pays $2 for each crate of cherry
tomato plants and $5 for each crate of Big Boy tomato plants. The daily demand equations for each type
of tomato plant are as follows:
x  35  10 p  8q demand for a crate of cherry tomatoes
y  90  17 p - 6q demand for a crate of Big Boy tomatoes
where p is the selling price for each crate of cherry tomatoes and q is the selling price for each crate of
Big Boy tomatoes. What price should the store sell each crate of tomatoes if they want to maximize their
profits?