College Mathematics: Learning Worksheets Chapter 14
Name ________________________________ Date ______________ Class ____________
Goal: To work with functions that involve more than one variable
1. If 32
(, ) 4 2 5 15,fxy x y xfind (3, 4).f
22
(, ) 4 2 5 15
fxy x y x
=+++
2. If 85
(, ) ,
x
x
y
gxy
find (1,5).g
33 11
57 19
85
8(1) 5(5)
(1, 5) 7(1) 10(5)
825
(1, 5) 750
(1, 5)
xy
x
y
g
g
g
+
+
=+
+
=+
==
Section 14-1 Functions of Several Variables
Definition:
An equation of the form (, )zfxydescribes a function of two independent variables
if, for each permissible ordered pair (x, y), there is one and only one value of z determined by
(, ).
f
xy
Examples of such functions are
(, )Axy xy Area of a rectangle
3. If 323
(, ,) 5 2 4 5,fxyz x y zfind (0,2,4).f
323
(, ,) 5 2 4 5
fxyz x y z
=+ −+
4. Find
(2000, 0.03,10)Afor (,,) .
A
Prt P Prt
(,,)
APrt P Prt
5. Find (12,8, 4)vfor ( , , ) .vlwh lwh
(, , )
vlwh lwh
=
In Problems 6 and 7, find the indicated function f of a single variable.
6. ( ) (5, )
f
yGy=for 32
(, ) 2 4 3.Gxy x xy y
32
32
() (,) 2 4 3
fy Gxy x xy y
==+−+
399
7. () (,3)
f
xNxxfor 22
(, ) 7 2 7.Nxy x y xy x
22
() (, ) 7 2 7
fx Nxy x y xy x
8. The Cobb-Douglas production function for a company is given by
0.3 0.7
(, ) 50 ,
f
xy x y=
where x is the utilization of labor and y is the utilization of capital. If the
company uses 1500 units of labor and 1200 units of capital, how many units
will be produced by the company?
College Mathematics: Learning Worksheets Chapter 14
400
College Mathematics: Learning Worksheets Chapter 14
Name ________________________________ Date ______________ Class ____________
Goal: To find partial derivatives of functions of several variables
Section 14-2 Partial Derivatives
Definition: Partial Derivative
With Respect to x:
0
(, ) (, )
(, ) lim
xh
zfxyhfxy
fxy
xh
With Respect to y:
0
(, ) (, )
(, ) lim
yk
z f xy h f xy
fxy
yk
Definition: Second–Order Partial Derivatives
If ( , ),zfxythen
2
zz
xy xy
y
xyx
2
zz
y
4. Find the value of (0,3)
y
fif 32
( , ) 3 5 8 2 15.fxy xy xy x y
We will first find the partial derivative with respect to y. Therefore, we consider only
the y as a variable and all other variables as constants.
(0,3) 2
y
f
=
5. Find the value of (3,1)
x
fif 22
(, ) 5 7 2 8 6.fxy x xy xy y
We will first find the partial derivative with respect to x. Therefore, we consider only
the x as a variable and all other variables as constants.
22
(, ) 5 7 2 8 6
fxy x xy xy y
=− +−+
8. (, )
yx
f
xyif 7
(, ) (7 2)
f
xy x y
We will first find the partial derivative with respect to y. Therefore, we consider only
the y as a variable and all other variables as constants.
7
(, ) (7 2)
fxy x y
9. ( , )
xx
Nxy
32
(,)12 5 7 4917
xx
Nxy x xy y x y
=+−+−+
10. (4, 4)
y
S
222
(, ) ln 3
(4)
(4, 4) 6( 4)
x
y
Sxy x y ye
Se
11. ( , )
x
Sxy
x
x
x
12. (10,5)
xx
N
32
(,)12 5 7 4917
xx
Nxy x xy y x y
=+−+−+
13. A firm produces two types of toy trains each week, x of type A and y of type
B. The weekly revenue and cost functions (in dollars) are as follows:
22
( , ) 50 70 0.07 0.04 0.04
R
xy x y xy x y
( , ) 7 9 18, 000Cxy x y
Find the profit function and then find (800,1000)
x
Pand (800,1000).
y
P
College Mathematics: Learning Worksheets Chapter 14
Name ________________________________ Date ______________ Class ____________
Goal: To find local maxima and minima of functions of several variables
1. If (, ) 14 9 21,fxy x y=−+find ( , )
x
f
xyand (, ),
y
f
xy then explain, using
Theorem 1, why (, )
f
xyhas no local extremum.
(, ) 14 9 21
(, ) 14
x
fxy x y
fxy
=−+
=
(, ) 14 9 21
(, ) 9
y
fxy x y
fxy
=−+
=−
Section 14-3 Maxima and Minima
Theorem 1: Local Extrema and Partial Derivatives
Let (,)
f
ab be a local extremum (a local maximum or a local minimum) for the
function f. If both
x
f
and y
f
exist at (a, b), then ( , ) 0
x
faband (,) 0.
y
fab
Theorem 2: Second-Derivative Test for Local Extremum
If
1. ( , )zfxy
x
y
3. All second–order partial derivatives of f exist in some circular region
containing (a, b) as center.
Case 1: If 20AC Band A < 0, then ( , )
f
ab is a local maximum.
f
Case 3: If 20,AC Bthen f has a saddle point at (a, b).
Case 4: If 20,AC Bthe test fails.
College Mathematics: Learning Worksheets Chapter 14
408
In Problems 2–5, use theorem 2 to find the local extremum.
2. 22
(, ) 7 5 3 4 9
f
xy x y x y=−+−+
Based on the above partial derivatives, there is a critical value at the point 32
14 5
(,).−−
Now find the values for A, B, and C.
14 5
3. 22
(, ) 2 4 7 3 12fxy x y x y
22
(, ) 2 4 7 3 12
(, ) 4 7
x
fxy x y x y
fxy x
22
(, ) 2 4 7 3 12
(, ) 8 3
y
fxy x y x y
fxy y
Based on the above partial derivatives, there is a critical value at the point 73
48
(,).
Now find the values for A, B, and C.
2
32
AC B
Since the above value is positive and the value of A is positive, the function has a
minimum at the point 73
48
(,).
409
4. 2
(, ) 4 5 4 7 19fxy xy y x y
2
(, ) 4 5 4 7 19
fxy xy y x y
2
(, ) 4 5 4 7 19
fxy xy y x y
Based on the above partial derivatives, there is a critical value at the point 3
4
(,1).
Now find the values for A, B, and C.
College Mathematics: Learning Worksheets Chapter 14
410
5. 42
1
2
(, ) 4 3fxy x y xy
2
3
(, ) 2 4
x
fxy x y
2
(, ) 2 4
y
fxy y x
Setting both equations equal to zero and solving will yield the critical points of (0, 0),
(2, 4) and (–2, –4). Now find the equations for A, B, and C.
2
6
xx
Ax
4
xy
B
2
yy
C
For the critical point (0, 0), the value of A will be 0. Therefore,
2
16
AC B
Since the above value is negative, the function has a saddle point at the point (0, 0).
For the critical point (2, 4), the value of A will be 24. Therefore,
2
32
AC B
Since the above value is positive and the value of A is positive, the function has a
minimum at the point (2, 4).
For the critical point (–2, –4), the value of A will be 24. Therefore,
2
32
AC B
Since the above value is positive and the value of A is positive, the function has a
minimum at the point (–2, –4).
411
6. A company produces two types of toy cars per day: x units of type A and y
units of type B. If the revenue and cost equations for the day are (in thousands
of dollars)
(, ) 25 31
R
xy x y=+
determine how many of each type of toy car should be produced per day to
maximize profit? What is the maximum profit?
(, ) (, ) (, )
Pxy Rxy Cxy
=−
Setting both equations equal to zero and solving will yield the critical point of
(50, 34). Now find the values for A, B, and C.
(,)
xx
APab
(,)
xy
B
Pab
(,)
yy
CPab
2
7
AC B
Since the above value is positive and the value of A is negative, the function has a
maximum at the point (50, 34). Therefore, you should produce 50 type A toy cars and 34 type
B toy cars to maximize your profit. The maximum profit will be
22
( , ) 2 5 4 30 22 7
Pxy x xy y x y
=− + − + + −
412
7. A rectangular box with no top and two parallel partitions must hold a volume
of 125 cubic inches. Find the dimensions that will require the least amount of
material.
Based on the information in the problem, we will let x = length, y = width, and z =
height. Therefore, 2 4
M
xy xz yz and the volume is 125.Vxyz Solve the
volume equation for z and reduce the original equation down to two variables yields
Setting both equations equal to zero and solving will yield the critical point of (10, 5).
Now find the values for A, B, and C.
1000
(,)
xx
AM ab
(,)
xy
B
Pab
(,)
yy
CPab
2
3
AC B
Since the above value is positive and the value of A is positive, the function has a
College Mathematics: Learning Worksheets Chapter 14
413
Name ________________________________ Date ______________ Class ____________
Goal: To find maxima and minima of functions of several variables using Lagrange
multipliers
Section 14-4 Maxima and Minima Using
Lagrange Multipliers
Theorem 1: Method of Lagrange Multipliers for Functions of Two Variables
Any local maxima or minima of the function ( , )zfxysubject to the constraint
(, ) 0gxy will be among those points 00
(, )
x
yfor which 000
(, , )xy
λ
is a solution of
the system
(,,)0
x
Fxy
λ
Procedure: Method of Lagrange Multipliers: Key Steps
Step 1: Write the problem in the form
Maximize (or minimize) ( , )zfxy
subject to ( , ) 0gxy
Step 2: Form the function F:
(, , ) (, ) (, )Fxy f xy gxy
λλ
Step 3: Find the critical points of F; that is, solve the system
λ
Step 4: If 000
(, , )xy
λ
is the only critical point of F, we assume that 00
(, )
x
ywill
always produce the solution to the problems we consider. If F has more than
one critical point, we evaluate (, )zfxyat 00
(, )
x
yfor each critical point
(, , )xy
λ
of F. For the problems we consider , we assume that the largest of
College Mathematics: Learning Worksheets Chapter 14
414
1. Use Theorem 1 to explain why no minima exist.
The problem will need to be rewritten in standard form first.
Now find the function F using the definition.
(, , ) (, ) (, )
Fxy f xy gxy
λλ
=+
Find the three partial derivatives.
(, , ) 10 7
x
Fxy
λλ
=+
x
y
Theorem 2: Method of Lagrange Multipliers for Functions of Three Variables
Any local maxima or minima of the function ( , , )wfxyzsubject to the constraint
(, ,) 0gxyz will be among the set of points 000
(, ,)
x
yz for which 0000
(, ,, )xyz
λ
is a
solution of the system
(,,,)0
x
Fxyz
λ
College Mathematics: Learning Worksheets Chapter 14
In Problems 2–5, use the method of Lagrange multipliers.
2. Maximize: ( , ) 9
f
xy xy
subject to: 2 2 6xy
Step 1
: Rewrite the problem.
f
Step 2
: Form the function F.
(, , ) (9 ) (2 2 6)
(, , ) 9 2 2 6
Fxy xy x y
Fxy xy x y
Step 3
: Find the critical points.
(, , ) 9 2 0
(, , ) 2 2 6 0
x
y
Fxy x
Fxy x y
Therefore,
2
9
y
and 2
9.x
2260
xy
2(6.75)
9
()
y
2(6.75)
9
()
x
416
3. Minimize: 22
(, ) 2 2
f
xy x y
subject to: 45 20xy
Step 1: Rewrite the problem.
f
Step 2: Form the function F.
(, , ) (, ) (, )
Fxy f xy gxy
λλ
Step 3: Find the critical points.
(, , ) 4 4 0
x
Fxy x
λλ
Therefore,
x
λ
and 5
4.y
λ
45200
xy
x
λ
5
4
y
λ
Step 4: Since the solution 80 100 80
,,
41 41 41
is the only critical point of F,