chapter 7 functions of several variables

© 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 1 of 62

Chapter 7

Functions of Several Variables


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


§ 7.1

Examples of Functions of Several Variables


Functions of More Than One Variable

Cost of Material

Tax and Homeowner Exemption

Level Curves

Section Outline



Definition Example

Function of Several Variables: A function that has more than one independent variable

wz

xywzyxf

5,,,

2



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Let . Compute g(1, 1) and g(0, -1).

22 2, yxyxg

31211211,1 22 g

21201201,0 22 g


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)


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


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


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.


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.


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.


§ 7.2

Partial Derivatives


Partial Derivatives

Computing Partial Derivatives

Evaluating Partial Derivatives at a Point

Local Approximation of f (x, y)

Demand Equations

Second Partial Derivative

Section Outline


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



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



This is the given function. yexyxf x ln, 32

Differentiate ln y.

CONTINUECONTINUEDD

yex

y

f x 132



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


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



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


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


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


§ 7.3

Maxima and Minima of Functions of Several Variables


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


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.


First-Derivative Test

If one or both of the partial derivatives does not exist, then there is no relative maximum or relative minimum.


Second-Derivative Test


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



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



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



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



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



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



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



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


§ 7.4

Lagrange Multipliers and Constrained Optimization


Background and Steps for Lagrange Multipliers

Using Lagrange Multipliers

Lagrange Multipliers in Application

Section Outline


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 ,,,,


Steps For Lagrange Multipliers

L-1

L-2

L-3



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



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



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



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’.



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



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



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


§ 7.5

The Method of Least Squares


Least Squares Error

Least Squares Line (Regression Line)

Determining a Least Squares Line

Section Outline


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


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.




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.



(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



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



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.


§ 7.6

Double Integrals


Double Integral of f (x, y) over a Region R

Evaluating Double Integrals

Double Integrals in “Application”

Section Outline


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))



The Double Integral


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.



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



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

chapter 7 functions of several variables

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