1. Chapter 7 Functions of Several Variables

2. Chapter Outline 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

3. § 7.1
Examples of Functions of Several Variables

4. Section Outline Functions of More Than One Variable Cost of Material
Tax and Homeowner Exemption
Level Curves

5. Functions of More Than One Variable
Definition
Example
Function of Several Variables: A function that has more than one independent variable

6. Functions of More Than One Variable
EXAMPLE
Let Compute g(1, 1) and g(0, -1).
SOLUTION

7. Cost of Material
EXAMPLE
(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.
SOLUTION
TOP
LEFT SIDE
RIGHT SIDE
BACK 3 5 xy yz xz
Cost (per sq ft)
Area (sq ft)

8. Cost of Material
CONTINUED
The total cost is the sum of the amount of cost for each side of the enclosure,
Similarly, the cost of the top is 3xy. Continuing in this way, we see that the total cost is

9. Tax & Homeowner Exemption
EXAMPLE
(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.

10. Tax & Homeowner Exemption
CONTINUED
SOLUTION
We are looking for T. We know that v = 200,000, x = 5000 and r = Therefore, we get
So, the real estate tax on the property with the given characteristics is $1875.

11. 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.

12. Level Curves
EXAMPLE
Find a function f (x, y) that has the curve y = 2/x2 as a level curve.
SOLUTION
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.
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.

13. § 7.2
Partial Derivatives

14. Section Outline Partial Derivatives Computing Partial Derivatives
Evaluating Partial Derivatives at a Point
Local Approximation of f (x, y)
Demand Equations
Second Partial Derivative

15. Definition: Partial Derivatives

16. Computing Partial Derivatives
EXAMPLE
Compute for
SOLUTION
To compute , we only differentiate factors (or terms) that contain x and
we interpret y to be a constant.
This is the given function.
Use the product rule where f (x) = x2 and g(x) = e3x.
To compute , we only differentiate factors (or terms) that contain y and
we interpret x to be a constant.

17. Computing Partial Derivatives
CONTINUED
This is the given function.
Differentiate ln y.

18. Computing Partial Derivatives
EXAMPLE
Compute for
SOLUTION
To compute , we treat every variable other than L as a constant. Therefore
This is the given function.
Rewrite as an exponent.
Bring exponent inside parentheses.
Note that K is a constant.
Differentiate.

19. Evaluating Partial Derivatives at a Point
EXAMPLE
Let Evaluate at (x, y, z) = (2, -1, 3).
SOLUTION

20. Partial Derivative Rule

21. Local Approximation of f (x, y)
We can generalize the interpretations of to yield the following general fact:
Partial derivatives can be computed for functions of any number of variables. When taking the partial derivative with respect to one variable, we treat the other variables as constant.

22. Local Approximation of f (x, y)
EXAMPLE
Let Interpret the result
SOLUTION
We showed in the last example that
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,

23. Demand Equations
EXAMPLE
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
SOLUTION
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).
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).

24. Second Partial Derivative
EXAMPLE
Let Find
SOLUTION
We first note that This means that to compute , we
must take the partial derivative of with respect to x.

25. § 7.3
Maxima and Minima of Functions of Several Variables

26. Section Outline 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

27. 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 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 near b.
Examples are forthcoming.

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

29. Second-Derivative Test

30. Finding Relative Maxima & Minima
EXAMPLE
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.
SOLUTION
We first use the first-derivative test.

31. Finding Relative Maxima & Minima
CONTINUED
Now we set both partial derivatives equal to 0 and then solve each for y.
Now we may set the equations equal to each other and solve for x.

32. Finding Relative Maxima & Minima
CONTINUED
We now determine the corresponding value of y by replacing x with 1 in the equation y = x + 2.
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

33. Finding Relative Maxima & Minima
CONTINUED
Since , we know, by the second-derivative test,
that f (x, y) has a relative maximum at (1, 3).

34. Finding Relative Maxima & Minima
EXAMPLE
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.
SOLUTION
We first use the first-derivative test.

35. Finding Relative Maxima & Minima
CONTINUED
Now we set both partial derivatives equal to 0 and then solve each for y.
Now we may set the equations equal to each other and solve for x.

36. Finding Relative Maxima & Minima
CONTINUED
We now determine the corresponding value of y by replacing x with 443 in the equation y = -0.1x
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

37. Finding Relative Maxima & Minima
CONTINUED
Since , we know, by the second-derivative test,
that R(x, y) has a relative maximum at (443, 236).

38. First Derivative Test – 3 Variables

39. § 7.4
Lagrange Multipliers and Constrained Optimization

40. Section Outline Background and Steps for Lagrange Multipliers
Using Lagrange Multipliers
Lagrange Multipliers in Application

41. Optimization Method of Lagrange multipliers:
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.

42. Steps For Lagrange Multipliers

43. Using Lagrange Multipliers
EXAMPLE
Maximize the function , subject to the constraint
SOLUTION
We have and
The equations L-1 to L-3, in this case, are

44. Using Lagrange Multipliers
CONTINUED
From the first two equations we see that
Therefore,
Substituting this expression for x into the third equation, we derive

45. Using Lagrange Multipliers
CONTINUED
Using y = 3/5, we find that
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

46. Lagrange Multipliers in Application
EXAMPLE
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.
SOLUTION
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
We can simplify this constraint equation as follows.
We must now determine the objective function. Since we wish to maximize area, our objective function should be about the quantity ‘area’.

47. Lagrange Multipliers in Application
CONTINUED
The area of the rectangular garden is xy. Therefore, our objective equation is
Therefore,
Now we calculate L-1, L-2, and L-3.

48. Lagrange Multipliers in Application
CONTINUED
From the first two equations we see that
Therefore,
Substituting this expression for y into the third equation, we derive

49. Lagrange Multipliers in Application
CONTINUED
Using x = 12, we find that
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

50. § 7.5
The Method of Least Squares

51. Section Outline Least Squares Error
Least Squares Line (Regression Line)
Determining a Least Squares Line

52. Example is forthcoming.
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.

53. Least Squares Line (Regression Line)

54. Determining a Least Squares Line
EXAMPLE
The following table gives the number in thousands of car-accident-related deaths in the U.S. for certain years.
(a) Find the straight line that provides the best least-squares fit to these data.
(b) Use the straight line found in part (a) to estimate the number of car-accident related deaths in (It is interesting that, while the number of drivers is obviously increasing with time, the number of car-accident-related deaths is actually decreasing, maybe because of improvements in car-manufacturing technologies and added safety measures.)

55. Determining a Least Squares Line
CONTINUED
SOLUTION
(a) The points are plotted in the figure below, where x denotes the number of years since The sums are calculated in the table in the following slide and then used to determine the values of A and B.

56. Determining a Least Squares Line
CONTINUED

57. Determining a Least Squares Line
CONTINUED
Therefore, the equation of the least-squares line is y = -.39x
(b) We use the straight line to estimate the number of car-accident-related deaths in 2012 by setting x = 22. Then we get
y = -.39(22) =
Therefore, we estimate the number of number of car-related accidental deaths to be thousand in 2012.

58. § 7.6
Double Integrals

59. Section Outline Double Integral of f (x, y) over a Region R
Evaluating Double Integrals
Double Integrals in “Application”

60. Example is forthcoming.
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.

61. The Double Integral

62. Evaluating Double Integrals
EXAMPLE
Calculate the iterated integral.
SOLUTION
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).
Now we carry out the integration with respect to x.
So the value of the iterated integral is 27/2.

63. Double Integrals in “Application”
EXAMPLE
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.
SOLUTION
The desired volume is given by the double integral By the
result just cited, this double integral is equal to the iterated integral
We first evaluate the inner integral.

64. Double Integrals in “Application”
CONTINUED
Now we carry out the integration with respect to y.
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.