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Chapter 7
Functions of Several Variables
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Examples of Functions of Several Variables
Partial Derivatives
Maxima and Minima of Functions of Several Variables
Lagrange Multipliers and Constrained Optimization
The Method of Least Squares
Double Integrals
Chapter Outline
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§ 7.1
Examples of Functions of Several Variables
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Functions of More Than One Variable
Cost of Material
Tax and Homeowner Exemption
Level Curves
Section Outline
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Functions of More Than One Variable
Definition Example
Function of Several Variables: A function that has more than one independent variable
wz
xywzyxf
5,,,
2
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Functions of More Than One Variable
EXAMPLEEXAMPLE
SOLUTIONSOLUTION
Let . Compute g(1, 1) and g(0, -1).
22 2, yxyxg
31211211,1 22 g
21201201,0 22 g
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Cost of Material
EXAMPLEEXAMPLE
SOLUTIONSOLUTION
(Cost) Find a formula C(x, y, z) that gives the cost of material for the rectangular enclose in the figure, with dimensions in feet, assuming that the material for the top costs $3 per square foot and the material for the back and two sides costs $5 per square foot.
TOP LEFT SIDE RIGHT SIDE BACK
3 5 5 5
xy yz yz xzArea (sq ft)
Cost (per sq ft)
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Cost of Material
The total cost is the sum of the amount of cost for each side of the enclosure,
CONTINUECONTINUEDD
.5feet squarein sideback of areaback offoot squareper cost xz
Similarly, the cost of the top is 3xy. Continuing in this way, we see that the total cost is
.51035553,, xzyzxyxzyzyzxyzyxC
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Tax & Homeowner Exemption
EXAMPLEEXAMPLE
(Tax and Homeowner Exemption) The value of residential property for tax purposes is usually much lower than its actual market value. If v is the market value, then the assessed value for real estate taxes might be only 40% of v. Suppose the property tax, T, in a community is given by the function
where v is the estimated market value of a property (in dollars), x is a homeowner’s exemption (a number of dollars depending on the type of property), and r is the tax rate (stated in dollars per hundred dollars) of net assessed value.
Determine the real estate tax on a property valued at $200,000 with a homeowner’s exemption of $5000, assuming a tax rate of $2.50 per hundred dollars of net assessed value.
, 4.0100
,, xvr
xvrfT
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Tax & Homeowner Exemption
SOLUTIONSOLUTION
We are looking for T. We know that v = 200,000, x = 5000 and r = 2.50. Therefore, we get
.18755000000,2004.0100
5.25000,000,200,5.2 fT
CONTINUECONTINUEDD
So, the real estate tax on the property with the given characteristics is $1875.
© 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 11 of 62
Level Curves
Definition Example
Level Curves: For a function f (x, y), a family of curves with equations f (x, y) = c where c is any constant
An example immediately follows.
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Level Curves
EXAMPLEEXAMPLE
SOLUTIONSOLUTION
Find a function f (x, y) that has the curve y = 2/x2 as a level curve.
2/2 xy
Since level curves occur where f (x, y) = c, then we must rewrite y = 2/x2 in that form.
This is the given equation of the level curve.
0/2 2 xy Subtract 2/x2 from both sides so that the left side resembles a function of the form f (x, y).
Therefore, we can say that y – 2/x2 = 0 is of the form f (x, y) = c, where c = 0. So, f (x, y) = y – 2/x2.
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§ 7.2
Partial Derivatives
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Partial Derivatives
Computing Partial Derivatives
Evaluating Partial Derivatives at a Point
Local Approximation of f (x, y)
Demand Equations
Second Partial Derivative
Section Outline
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Partial Derivatives
Definition Example
Partial Derivative of f (x, y)
with respect to x: Written ,
the derivative of f (x, y), where y is treated as a constant and f (x, y) is considered as a function of x alone
If , then
x
f
432, yxyxf
.8
and 6
33
42
yxy
f
yxx
f
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Computing Partial Derivatives
EXAMPLEEXAMPLE
SOLUTIONSOLUTION
Compute for
To compute , we only differentiate factors (or terms) that contain x and
we interpret y to be a constant.
This is the given function.
.ln, 32 yexyxf x
x
f
yexyxf x ln, 32
Use the product rule where f (x) = x2 and g(x) = e3x.
233 32ln xeexyx
f xx
y
f
x
f
and
To compute , we only differentiate factors (or terms) that contain y and
we interpret x to be a constant.y
f
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Computing Partial Derivatives
This is the given function. yexyxf x ln, 32
Differentiate ln y.
CONTINUECONTINUEDD
yex
y
f x 132
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Computing Partial Derivatives
EXAMPLEEXAMPLE
SOLUTIONSOLUTION
Compute for
To compute , we treat every variable other than L as a constant. Therefore
This is the given function.
.3, LKKLf
L
f
L
f
LKKLf 3,
Rewrite as an exponent. 213, LKKLf
Bring exponent inside parentheses.
21213, KLKLf
Note that K is a constant. 21213, LKKLf
Differentiate. L
KLK
L
f
2
33
2
1 2121
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Evaluating Partial Derivatives at a Point
EXAMPLEEXAMPLE
SOLUTIONSOLUTION
Let Evaluate at (x, y, z) = (2, -1, 3).
.5,, 2 zxyzyxfy
f
.1231223,1,2,2
y
fxyz
y
f
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Local Approximation of f (x, y)
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Local Approximation of f (x, y)
EXAMPLEEXAMPLE
SOLUTIONSOLUTION
Let Interpret the result
.5,, 2 zxyzyxf .123,1,2 y
f
We showed in the last example that .123,1,2
y
f
This means that if x and z are kept constant and y is allowed to vary near -1, then f (x, y, z) changes at a rate 12 times the change in y (but in a negative direction). That is, if y increases by one small unit, then f (x, y, z) decreases by approximately 12 units. If y increases by h units (where h is small), then f (x, y, z) decreases by approximately 12h. That is,
.123,1,23,1,2 hfhf
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Demand Equations
EXAMPLEEXAMPLE
SOLUTIONSOLUTION
The demand for a certain gas-guzzling car is given by f (p1, p2), where p1 is
the price of the car and p2 is the price of gasoline. Explain why and 01
p
f
is the rate at which demand for the car changes as the price of the car
changes. This partial derivative is always less than zero since, as the price of the car increases, the demand for the car will decrease (and visa versa).
.02
p
f
1p
f
is the rate at which demand for the car changes as the price of gasoline
changes. This partial derivative is always less than zero since, as the price of gasoline increases, the demand for the car will decrease (and visa versa).
2p
f
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Second Partial Derivative
EXAMPLEEXAMPLE
SOLUTIONSOLUTION
Let . Find .2
yx
f
We first note that This means that to compute , we
must take the partial derivative of with respect to x.
34, yyxxeyxf y
. 2
y
f
xyx
f
yx
f
2
y
f
3242
43 xeyxxexy
f
xyx
f yy
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§ 7.3
Maxima and Minima of Functions of Several Variables
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Relative Maxima and Minima
First Derivative Test for Functions of Two Variables
Second Derivative Test for Functions of Two Variables
Finding Relative Maxima and Minima
Section Outline
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Relative Maxima & Minima
Definition Example
Relative Maximum of f (x, y): f (x, y) has a relative maximum when x = a, y = b if f (x, y) is at most equal to f (a, b) whenever x is near a and y is is near b.
Examples are forthcoming.
Definition Example
Relative Minimum of f (x, y): f (x, y) has a relative minimum when x = a, y = b if f (x, y) is at least equal to f (a, b) whenever x is near a and y is is near b.
Examples are forthcoming.
© 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 27 of 62
First-Derivative Test
If one or both of the partial derivatives does not exist, then there is no relative maximum or relative minimum.
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Second-Derivative Test
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Finding Relative Maxima & Minima
EXAMPLEEXAMPLE
SOLUTIONSOLUTION
Find all points (x, y) where f (x, y) has a possible relative maximum or minimum. Then use the second-derivative test to determine, if possible, the nature of f (x, y) at each of these points. If the second-derivative test is inconclusive, so state.
2216432, 22 yxyxyxyxf
We first use the first-derivative test.
422
yxx
f
1662
yxy
f
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Finding Relative Maxima & Minima
Now we set both partial derivatives equal to 0 and then solve each for y.
0422 yx 01662 yx
CONTINUECONTINUEDD
2xy3
8
3
1 xy
Now we may set the equations equal to each other and solve for x.
3
8
3
12 xx
863 xx
862 x
22 x
1x
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Finding Relative Maxima & Minima
We now determine the corresponding value of y by replacing x with 1 in the equation y = x + 2.
CONTINUECONTINUEDD
321 y
So we now know that if there is a relative maximum or minimum for the function, it occurs at (1, 3). To determine more about this point, we employ the second-derivative test. To do so, we must first calculate
.,22
2
2
2
2
yx
f
y
f
x
fyxD
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Finding Relative Maxima & Minima
Since , we know, by the second-derivative test,
that f (x, y) has a relative maximum at (1, 3).
CONTINUECONTINUEDD
0 and 0,2
yx
fyxD
24222
2
yxxx
f
xx
f
616622
2
yxyy
f
yy
f
216622
yx
xy
f
xyx
f
8262, 2
22
2
2
2
2
yx
f
y
f
x
fyxD
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Finding Relative Maxima & Minima
EXAMPLEEXAMPLE
SOLUTIONSOLUTION
A monopolist manufactures and sells two competing products, call them I and II, that cost $30 and $20 per unit, respectively, to produce. The revenue from marketing x units of product I and y units of product II is
Find the values of x and y that maximize the monopolist’s profits.
.2.01.004.011298, 22 yxxyyxyxR
We first use the first-derivative test.
xyx
R2.004.098
yxy
R4.004.0112
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Finding Relative Maxima & Minima
Now we set both partial derivatives equal to 0 and then solve each for y.
CONTINUECONTINUEDD
Now we may set the equations equal to each other and solve for x.
02.004.098 xy 04.004.0112 yx
24505 xy 2801.0 xy
2801.024505 xx
28024509.4 x
21709.4 x
443x
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Finding Relative Maxima & Minima
We now determine the corresponding value of y by replacing x with 443 in the equation y = -0.1x + 280.
CONTINUECONTINUEDD
2362804431.0 y
So we now know that revenue is maximized at the point (443, 236). Let’s verify this using the second-derivative test. To do so, we must first calculate
.,22
2
2
2
2
yx
R
y
R
x
RyxD
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Finding Relative Maxima & Minima
Since , we know, by the second-derivative test,
that R(x, y) has a relative maximum at (443, 236).
CONTINUECONTINUEDD
0 and 0,2
yx
RyxD
2.02.004.0982
2
xyxx
R
xx
R
4.04.004.01122
2
yxyy
R
yy
R
04.04.004.01122
yx
xy
R
xyx
R
0784.004.04.02.0, 2
22
2
2
2
2
yx
R
y
R
x
RyxD
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§ 7.4
Lagrange Multipliers and Constrained Optimization
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Background and Steps for Lagrange Multipliers
Using Lagrange Multipliers
Lagrange Multipliers in Application
Section Outline
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Optimization
In this section, we will optimize an objective equation f (x, y) given a constraint equation g(x, y). However, the methods of chapter 2 will not work, so we must do something different. Therefore we must use the following equation and theorem.
yxgyxfyxF ,,,,
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Steps For Lagrange Multipliers
L-1
L-2
L-3
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Using Lagrange Multipliers
EXAMPLEEXAMPLE
SOLUTIONSOLUTION
Maximize the function , subject to the constraint
22 yx
We have and
. 32,, 22 yxyxyxF
022 x
x
F
.032 yx
32, ,, 22 yxyxgyxyxf
The equations L-1 to L-3, in this case, are
02 y
y
F
.032
yxF
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Using Lagrange Multipliers
From the first two equations we see that
CONTINUECONTINUEDD
.2yx
Therefore,
.2yx
Substituting this expression for x into the third equation, we derive
032 yx
0322 yy
035 y
5
3y
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Using Lagrange Multipliers
Using y = 3/5, we find that
CONTINUECONTINUEDD
.5
6
5
32
5
6
5
32
x
So the maximum value of x2 + y2 with x and y subject to the constraint occurs when x = 6/5, y = 3/5, and That maximum value is.5/6
.8.125
45
25
9
25
36
5
3
5
622
© 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 44 of 62
Lagrange Multipliers in Application
EXAMPLEEXAMPLE
SOLUTIONSOLUTION
Four hundred eighty dollars are available to fence in a rectangular garden. The fencing for the north and south sides of the garden costs $10 per foot and the fencing for the east and west sides costs $15 per foot. Find the dimensions of the largest possible garden.
Let x represent the length of the garden on the north and south sides and y represent the east and west sides. Since we want to use all $480, we know that
.48015151010 yyxx
We can simplify this constraint equation as follows.
04803020, yxyxg
We must now determine the objective function. Since we wish to maximize area, our objective function should be about the quantity ‘area’.
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Lagrange Multipliers in Application
The area of the rectangular garden is xy. Therefore, our objective equation is
., xyyxfA
Therefore,
CONTINUECONTINUEDD
. 4803020,,,, yxxyyxgyxfyxF
Now we calculate L-1, L-2, and L-3.
020 y
x
F
030 x
y
F
04803020
yxF
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Lagrange Multipliers in Application
From the first two equations we see that
CONTINUECONTINUEDD
.3020
xy
Therefore,
.3
2xy
Substituting this expression for y into the third equation, we derive
04803020 yx
04803
23020
xx
04802020 xx
12x
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Lagrange Multipliers in Application
Using x = 12, we find that
CONTINUECONTINUEDD
.5
2
30
12
8123
2
y
So the maximum value of xy with x and y subject to the constraint occurs when x = 12, y = 8, and That maximum value is.5/2
.feet square 96812
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§ 7.5
The Method of Least Squares
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Least Squares Error
Least Squares Line (Regression Line)
Determining a Least Squares Line
Section Outline
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Least Squares Error
Definition Example
Least Squares Error: The total error in approximating the data points (x1, y1),...., (xN, yN) by a line y = Ax + B, measured by the sum E of the squares of the vertical distances from the points to the line,
Example is forthcoming.
222
21 NEEEE
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Least Squares Line (Regression Line)
Definition Example
Least Squares Line (Regression Line): A straight line y = Ax + B for which the error E is as small as possible.
Example is forthcoming.
© 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 52 of 62
Determining a Least Squares Line
EXAMPLEEXAMPLE
Table 5 shows the 1994 price of a gallon (in U.S. dollars) of fuel and the average miles driven per automobile for several countries.
(a) Find the straight line that provides the best least-squares fit to these data.(b) In 1994, the price of gas in Japan was $4.14 per gallon. Use the straight line of part (a) to estimate the average number of miles automobiles were driven in Japan.
© 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 53 of 62
Determining a Least Squares Line
(a) The points are plotted in the figure below. The sums are calculated in the table below and then used to determine the values of A and B.
CONTINUECONTINUEDD
SOLUTIONSOLUTION
0
2000
4000
6000
8000
10000
12000
0 1 2 3 4
Price per Gallon
Ave
rag
e M
iles
per
Au
to
© 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 54 of 62
Determining a Least Squares Line
CONTINUECONTINUEDD
x y xy x2
1.57 10,371 16,282.47 2.4649
2.86 10,186 29,131.96 8.1796
3.31 8740 28,929.4 10.9561
3.34 7674 25,631.16 11.1556
3.44 7456 25,648.64 11.8336
1.24 11,099 13,762.76 1.5376
∑ x = 15.76 ∑ y = 55,526 ∑ xy = 139,386.4 ∑ x2 = 46.1274
© 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 55 of 62
Determining a Least Squares Line
CONTINUECONTINUEDD
824.1365
76.151274.466
526,5576.154.386,13962
A
898.841,12
6
76.15824.1365526,55
B
Therefore, the equation of the least-squares line is y = -1365.824x + 12,841.898.
(b) We use the straight line to estimate the average number of miles automobiles were driven in Japan in 1994 by setting x = 4.14. Then we get
y = -1365.824(4.14) + 12,841.898 ≈ 7187.
Therefore, we estimate the average number of miles per auto in Japan in 1994 to be 7187.
© 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 56 of 62
§ 7.6
Double Integrals
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Double Integral of f (x, y) over a Region R
Evaluating Double Integrals
Double Integrals in “Application”
Section Outline
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Double Integral of f (x, y) over a Region R
Definition Example
Double Integral of f (x, y) over a Region R: For a given function f (x, y) and a region R in the xy-plane, the volume of the solid above the region (given by the graph of f (x, y)) minus the volume of the solid below the region (given by the graph of f (x, y))
Example is forthcoming.
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The Double Integral
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Evaluating Double Integrals
EXAMPLEEXAMPLE
SOLUTIONSOLUTION
Calculate the iterated integral.
Here g(x) = x and h(x) = 2x. We evaluate the inner integral first. The variable in this integral is y (because of the dy).
3
0
2dxydy
x
x
22222
2
2
3
22
2
2x
xxyydy
x
x
x
x
Now we carry out the integration with respect to x.
2
270
2
13
2
1
2
1
2
3 333
0
33
0
2 xdxx
So the value of the iterated integral is 27/2.
© 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 61 of 62
Double Integrals in “Application”
EXAMPLEEXAMPLE
SOLUTIONSOLUTION
Calculate the volume over the following region R bounded above by the graph of f (x, y) = x2 + y2.
R is the rectangle bounded by the lines x = 1, x = 3, y = 0, and y = 1.
The desired volume is given by the double integral . By the
result just cited, this double integral is equal to the iterated integral
R
dxdyyx 22
1
0
3
1
22 .dydxyx
We first evaluate the inner integral.
22223
23
3
1
233
1
22 23
26
3
1391
3
13
3
3
3yyyyyxy
xdxyx
© 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 62 of 62
Double Integrals in “Application”
Now we carry out the integration with respect to y.
3
2800
3
2
3
260
3
20
3
261
3
21
3
26
3
2
3
262
3
26 331
0
31
0
2
yydyy
CONTINUECONTINUEDD
So the value of the iterated integral is 28/3.
Notice that we could have set up the initial double integral as follows.
This would have given us the same answer.
3
1
1
0
22 dxdyyx