8 Calculus of Several Variables Functions of Several Variables

8 Calculus of Several Variables Functions of Several Variables
Partial Derivatives Maxima and Minima of Functions of Several Variables The Method of Least Squares Constrained Maxima and Minima and the Method of Lagrange Multipliers Double Integrals

Functions of Several Variables
8.1 Functions of Several Variables

Functions of Two Variables
A real-valued function of two variables f, consists of A set A of ordered pairs of real numbers (x, y) called the domain of the function. A rule that associates with each ordered pair in the domain of f one and only one real number, denoted by z = f(x, y).

Examples Let f be the function defined by
Compute f(0, 0), f(1, 2), and f(2, 1). Solution The domain of a function of two variables f(x, y), is a set of ordered pairs of real numbers and may therefore be viewed as a subset of the xy-plane. Example 1, page 536

Examples Find the domain of the function Solution
f(x, y) is defined for all real values of x and y, so the domain of the function f is the set of all points (x, y) in the xy-plane. Example 2, page 536

g(x, y) is defined for all x ≠ y, so the domain of the function g is the set of all points (x, y) in the xy-plane except those lying on the y = x line. y y = x x Example 2, page 536

We require that 1 – x2 – y2  0 or x2 + y2  1 which is the set of all points (x, y) lying on and inside the circle of radius 1 with center at the origin: y 1 –1 x2 + y2 = 1 x –1 1 Example 2, page 536

Applied Example: Revenue Functions
Acrosonic manufactures a bookshelf loudspeaker system that may be bought fully assembled or in a kit. The demand equations that relate the unit price, p and q, to the quantities demanded weekly, x and y, of the assembled and kit versions of the loudspeaker systems are given by What is the weekly total revenue function R(x, y)? What is the domain of the function R? Applied Example 3, page 537

Solution The weekly revenue from selling x units assembled speaker systems at p dollars per unit is given by xp dollars. Similarly, the weekly revenue from selling y speaker kits at q dollars per unit is given by yq dollars. Therefore, the weekly total revenue function R is given by Applied Example 3, page 537

Solution To find the domain of the function R, note that the quantities x, y, p, and q must be nonnegative, which leads to the following system of linear inequalities: Thus, the graph of the domain is: y 2000 1000 D x Applied Example 3, page 537

Graphs of Functions of Two Variables
Consider the task of locating P(1, 2, 3) in 3-space: One method to achieve this is to start at the origin and measure out from there, axis by axis: z P(1, 2, 3) 3 y 1 2 x

Consider the task of locating P(1, 2, 3) in 3-space: Another common method is to find the xy coordinate and from there elevate to the level of the z value: z P(1, 2, 3) 3 y 2 1 (1, 2) x

Locate the following points in 3-space: Q(–1, 2, 3), R(1, 2, –2), and S(1, –1, 0). Solution z Q(–1, 2, 3) 3 y S(1, –1, 0) x –2 R(1, 2, –2)

The graph of a function in 3-space is a surface. For every (x, y) in the domain of f, there is a z value on the surface. z z = f(x, y) (x, y, z) y (x, y) x

Level Curves The graph of a function of two variables is often difficult to sketch. It can therefore be useful to apply the method used to construct topographic maps. This method is relatively easy to apply and conveys sufficient information to enable one to obtain a feel for the graph of the function.

Level Curves In the 3-space graph we just saw, we can delineate the contour of the graph as it is cut by a z = c plane: z z = f(x, y) z = c y f(x, y) = c x

Examples Sketch a contour map of the function f(x, y) = x2 + y2.
Solution The function f(x, y) = x2 + y2 is a revolving parabola called a paraboloid. z f(x, y) = x2 + y2 y Example 5, page 540 x

Examples Sketch a contour map of the function f(x, y) = x2 + y2.
Solution A level curve is the graph of the equation x2 + y2 = c, which describes a circle with radius Taking different values of c we obtain: z f(x, y) = x2 + y2 y x2 + y2 = 16 4 2 – 2 – 4 x2 + y2 = 9 x2 + y2 = 4 x2 + y2 = 16 x2 + y2 = 1 x2 + y2 = 0 x – – x2 + y2 = 9 x2 + y2 = 4 x2 + y2 = 1 y x2 + y2 = 0 Example 5, page 540 x

Examples Sketch level curves of the function f(x, y) = 2x2 – y corresponding to z = –2, –1, 0, 1, and 2. Solution The level curves are the graphs of the equation 2x2 – y = k or for k = –2, –1, 0, 1, and 2: y 2x2 – y = – 2 4 3 2 1 – 1 – 2 2x2 – y = – 1 2x2 – y = 0 2x2 – y = 1 2x2 – y = 2 x – 2 – Example 6, page 540

8.2 Partial Derivatives

First Partial Derivatives
First Partial Derivatives of f(x, y) Suppose f(x, y) is a function of two variables x and y. Then, the first partial derivative of f with respect to x at the point (x, y) is provided the limit exists. The first partial derivative of f with respect to y at the point (x, y) is

Geometric Interpretation of the Partial Derivative
z f(x, y) y x

z f(x, y) f(x, b) y = b plane b y a (a, b) x

z f(x, y) y x

z f(x, y) f(c, y) x = c plane d y c (c, d) x

Examples Find the partial derivatives ∂f/∂x and ∂f/∂y of the function
Use the partials to determine the rate of change of f in the x-direction and in the y-direction at the point (1, 2) . Solution To compute ∂f/∂x, think of the variable y as a constant and differentiate the resulting function of x with respect to x: Example 1, page 546

Use the partials to determine the rate of change of f in the x-direction and in the y-direction at the point (1, 2). Solution To compute ∂f/∂y, think of the variable x as a constant and differentiate the resulting function of y with respect to y: Example 1, page 546

Use the partials to determine the rate of change of f in the x-direction and in the y-direction at the point (1, 2). Solution The rate of change of f in the x-direction at the point (1, 2) is given by The rate of change of f in the y-direction at the point (1, 2) is given by Example 1, page 546

Examples Find the first partial derivatives of the function Solution
To compute ∂w/∂x, think of the variable y as a constant and differentiate the resulting function of x with respect to x: Example 2, page 547

To compute ∂w/∂y, think of the variable x as a constant and differentiate the resulting function of y with respect to y: Example 2, page 547

To compute ∂g/∂s, think of the variable t as a constant and differentiate the resulting function of s with respect to s: Example 2, page 547

To compute ∂g/∂t, think of the variable s as a constant and differentiate the resulting function of t with respect to t: Example 2, page 547

To compute ∂h/∂u, think of the variable v as a constant and differentiate the resulting function of u with respect to u: Example 2, page 547

To compute ∂h/∂v, think of the variable u as a constant and differentiate the resulting function of v with respect to v: Example 2, page 547

Here we have a function of three variables, x, y, and z, and we are required to compute For short, we can label these first partial derivatives respectively fx, fy, and fz. Example 3, page 549

To find fx, think of the variables y and z as a constant and differentiate the resulting function of x with respect to x: Example 3, page 549

To find fy, think of the variables x and z as a constant and differentiate the resulting function of y with respect to y: Example 3, page 549

To find fz, think of the variables x and y as a constant and differentiate the resulting function of z with respect to z: Example 3, page 549

The Cobb-Douglas Production Function
The Cobb-Douglass Production Function is of the form f(x, y) = axby1 – b (0 < b < 1) where a and b are positive constants, x stands for the cost of labor, y stands for the cost of capital equipment, and f measures the output of the finished product.

The Cobb-Douglas Production Function
The Cobb-Douglass Production Function is of the form f(x, y) = axby1 – b (0 < b < 1) The first partial derivative fx is called the marginal productivity of labor. It measures the rate of change of production with respect to the amount of money spent on labor, with the level of capital kept constant. The first partial derivative fy is called the marginal productivity of capital. It measures the rate of change of production with respect to the amount of money spent on capital, with the level of labor kept constant.

Applied Example: Marginal Productivity
A certain country’s production in the early years following World War II is described by the function f(x, y) = 30x2/3y1/3 when x units of labor and y units of capital were used. Compute fx and fy. Find the marginal productivity of labor and the marginal productivity of capital when the amount expended on labor and capital was 125 units and 27 units, respectively. Should the government have encouraged capital investment rather than increase expenditure on labor to increase the country’s productivity? Applied Example 4, page 550

f(x, y) = 30x2/3y1/3 Solution The first partial derivatives are Applied Example 4, page 550

f(x, y) = 30x2/3y1/3 Solution The required marginal productivity of labor is given by or 12 units of output per unit increase in labor expenditure (keeping capital constant). The required marginal productivity of capital is given by or 27 7/9 units of output per unit increase in capital expenditure (keeping labor constant). Applied Example 4, page 550

f(x, y) = 30x2/3y1/3 Solution The government should definitely have encouraged capital investment. A unit increase in capital expenditure resulted in a much faster increase in productivity than a unit increase in labor: 27 7/9 versus 12 per unit of investment, respectively. Applied Example 4, page 550

Second Order Partial Derivatives
The first partial derivatives fx(x, y) and fy(x, y) of a function f(x, y) of two variables x and y are also functions of x and y. As such, we may differentiate each of the functions fx and fy to obtain the second-order partial derivatives of f.

Differentiating the function fx with respect to x leads to the second partial derivative But the function fx can also be differentiated with respect to y leading to a different second partial derivative

Similarly, differentiating the function fy with respect to y leads to the second partial derivative Finally, the function fy can also be differentiated with respect to x leading to the second partial derivative

Thus, four second-order partial derivatives can be obtained of a function of two variables: When both are continuous

Examples Find the second-order partial derivatives of the function
Solution First, calculate fx and use it to find fxx and fxy: Example 6, page 552

Solution Then, calculate fy and use it to find fyx and fyy: Example 6, page 552

Solution First, calculate fx and use it to find fxx and fxy: Example 7, page 553

Solution Then, calculate fy and use it to find fyx and fyy: Example 7, page 553

Maxima and Minima of Functions of Several Variables
8.3 Maxima and Minima of Functions of Several Variables

Relative Extrema of a Function of Two Variables
Let f be a function defined on a region R containing the point (a, b). Then, f has a relative maximum at (a, b) if f(x, y)  f(a, b) for all points (x, y) that are sufficiently close to (a, b). The number f(a, b) is called a relative maximum value. Similarly, f has a relative minimum at (a, b) if f(x, y)  f(a, b) for all points (x, y) that are sufficiently close to (a, b). The number f(a, b) is called a relative minimum value.

Graphic Example There is a relative maximum at (a, b). z x y (a, b)

Graphic Example There is an absolute maximum at (c, d).
(It is also a relative maximum) z x y (c, d)

Graphic Example There is a relative minimum at (e, f ). z x y (e, f )

Graphic Example There is an absolute minimum at (g, h).
(It is also a relative minimum) z x (g, h) y

Relative Minima At a minimum point of the graph of a function of two variables, such as point (a, b) below, the plane tangent to the graph of the function is horizontal (assuming the surface of the graph is smooth): z y (a, b) x

Relative Minima Thus, at a minimum point, the graph of the function has a slope of zero along a direction parallel to the x-axis: z y (a, b) x

Relative Minima Similarly, at a minimum point, the graph of the function has a slope of zero along a direction parallel to the y-axis: z y (a, b) x

Relative Maxima At a maximum point of the graph of a function of two variables, such as point (a, b) below, the plane tangent to the graph of the function is horizontal (assuming the surface of the graph is smooth): z y (a, b) x

Relative Maxima Thus, at a maximum point, the graph of the function has a slope of zero along a direction parallel to the x-axis: z y (a, b) x

Relative Maxima Similarly, at a maximum point, the graph of the function has a slope of zero along a direction parallel to the y-axis: z y (a, b) x

Saddle Point In the case of a saddle point, both partials are equal to zero, but the point is neither a maximum nor a minimum. z y x

Saddle Point In the case of a saddle point, the function is at a minimum along one vertical plane… z y (a, b) x

Saddle Point … but at a maximum along the perpendicular vertical plane. z y (a, b) x

Extrema When Partial Derivatives are Not Defined
A maximum (or minimum) may also occur when both partial derivatives are not defined, such as point (a, b) in the graph below: z (a, b, f(a, b)) y (a, b) x

Critical Point of a Function
A critical point of f is a point (a, b) in the domain of f such that both or at least one of the partial derivatives does not exist.

Determining Relative Extrema
Find the critical points of f(x, y) by solving the system of simultaneous equations fx = 0 fy = 0 The second derivative test: Let D(x, y) = fxx fyy – f 2xy Then, a. D(a, b) > 0 and fxx(a, b) < 0 implies that f(x, y) has a relative maximum at the point (a, b). b. D(a, b) > 0 and fxx(a, b) > 0 implies that f(x, y) has a relative minimum at the point (a, b). c. D(a, b) < 0 implies that f(x, y) has neither a relative maximum nor a relative minimum at the point (a, b), it has instead a saddle point. d. D(a, b) = 0 implies that the test is inconclusive, so some other technique must be used to solve the problem.

Examples Find the relative extrema of the function Solution
We have fx = 2x and fy = 2y. To find the critical points, we set fx = 0 and fy = 0 and solve the resulting system of simultaneous equations 2x = 0 and y = 0 obtaining x = 0, y = 0, or (0, 0), as the sole critical point. Next, apply the second derivative test to determine the nature of the critical point (0, 0). We compute fxx = 2, fyy = 2, and fxy = 0, Thus, D(x, y) = fxx fyy – f 2xy = (2)(2) – (0)2 = 4. Example 1, page 561

We have D(x, y) = 4, and in particular, D(0, 0) = 4. Since D(0, 0) > 0 and fxx = 2 > 0, we conclude that f has a relative minimum at the point (0, 0). The relative minimum value, f (0, 0) = 0, also happens to be the absolute minimum of f. Example 1, page 561

The relative minimum value, f(0, 0) = 0, also happens to be the absolute minimum of f: Absolute minimum at (0, 0, 0). Example 1, page 561

Examples Find the relative extrema of the function Solution We have
To find the critical points, we set fx = 0 and fy = 0 and solve the resulting system of simultaneous equations 6x – 4y – 4 = and – 4x + 8y + 8 = 0 obtaining x = 0, y = –1, or (0, –1), as the sole critical point. Next, apply the second derivative test to determine the nature of the critical point (0, –1). We compute fxx = 6, fyy = 8, and fxy = – 4, Thus, D(x, y) = fxx · fyy – f 2xy = (6)(8) – (– 4)2 = 32. Example 2, page 562

We have D(x, y) = 32, and in particular, D(0, –1) = 32. Since D(0, –1) > 0 and fxx = 6 > 0, we conclude that f has a relative minimum at the point (0, –1). The relative minimum value, f (0, –1) = 0, also happens to be the absolute minimum of f. Example 2, page 562

D(x, y) = fxx · fyy – f 2xy = (2)· 24(y – 1) – (0)2 = 48(y – 1)
Examples Find the relative extrema of the function Solution We have To find the critical points, we set fx = 0 and fy = 0 and solve the resulting system of simultaneous equations The first equation implies that x = 0, while the second equation implies that y = –1 or y = 3. Thus, there are two critical points of f : (0, –1) and (0, 3). To apply the second derivative test, we calculate fxx = fyy = 24(y – 1) fxy = 0 D(x, y) = fxx · fyy – f 2xy = (2)· 24(y – 1) – (0)2 = 48(y – 1) Example 3, page 562

Apply the second derivative test to the critical point (0, –1): We have D(x, y) = 48(y – 1). In particular, D(0, –1) = 48[(–1) – 1] = – 96. Since D(0, –1) = – 96 < 0 we conclude that f has a saddle point at (0, –1). The saddle point value is f (0, –1) = 22, so there is a saddle point at (0, –1, 22). Example 3, page 562

Apply the second derivative test to the critical point (0, 3): We have D(x, y) = 48(y – 1). In particular, D(0, 3) = 48[(3) – 1] = 96. Since D(0, –1) = 96 > 0 and fxx (0, 3) = 2 > 0, we conclude that f has a relative minimum at the point (0, 3). The relative minimum value, f (0, 3) = –106, so there is a relative minimum at (0, 3, –106). Example 3, page 562

Applied Example: Maximizing Profit
The total weekly revenue that Acrosonic realizes in producing and selling its loudspeaker system is given by where x denotes the number of fully assembled units and y denotes the number of kits produced and sold each week. The total weekly cost attributable to the production of these loudspeakers is Determine how many assembled units and how many kits should be produced per week to maximize profits. Applied Example 3, page 563

Solution The contribution to Acrosonic’s weekly profit stemming from the production and sale of the bookshelf loudspeaker system is given by Applied Example 3, page 563

Solution We have To find the relative maximum of the profit function P, we first locate the critical points of P. Setting Px and Py equal to zero, we obtain Solving the system of equations we get x = 208 and y = 64. Therefore, P has only one critical point at (208, 64). Applied Example 3, page 563

Solution To test if the point (208, 64) is a solution to the problem, we use the second derivative test. We compute So, In particular, D(208, 64) = 5/16 > 0. Since D(208, 64) > 0 and Pxx(208, 64) < 0, the point (208, 64) yields a relative maximum of P. Applied Example 3, page 563

Solution The relative maximum at (208, 64) is also the absolute maximum of P. We conclude that Acrosonic can maximize its weekly profit by manufacturing 208 assembled units and 64 kits. The maximum weekly profit realizable with this output is Applied Example 3, page 563

The Method of Least Squares
8.4 The Method of Least Squares

Suppose we are given the data points P1(x1, y1), P2(x2, y2), P3(x3, y3), P4(x4, y4), and P5(x5, y5) that describe the relationship between two variables x and y. By plotting these data points, we obtain a scatter diagram: y P5 10 5 P3 P2 P4 P1 x 5 10

Suppose we try to fit a straight line L to the data points P1, P2, P3, P4, and P5. The line will miss these points by the amounts d1, d2, d3, d4, and d5 respectively. y L 10 5 x 5 10

The principle of least squares states that the straight line L that fits the data points best is the one chosen by requiring that the sum of the squares of d1, d2, d3, d4, and d5, that is be made as small as possible. y L 10 5 x 5 10

Suppose the regression line L is y = f(x) = mx + b, where m and b are to be determined. The distances d1, d2, d3, d4, and d5, represent the errors the line L is making in estimating these points, so that y L 10 5 x 5 10

Observe that This may be viewed as a function of two variables m and b. Thus, the least-squares criterion is equivalent to minimizing the function

We want to minimize We first find the partial derivative with respect to m:

We want to minimize We now find the partial derivative with respect to b:

Setting gives and Solving the two simultaneous equations for m and b then leads to an equation y = mx + b. This equation will be the ‘best fit’ line, or regression line for the given data points.

Suppose we are given n data points: P1(x1, y1), P2(x2, y2), P3(x3, y3), … , Pn(xn, yn) Then, the least-squares (regression) line for the data is given by the linear equation y = f(x) = mx + b where the constants m and b satisfy the equations and simultaneously. These last two equations are called normal equations.

Example Find the equation of the least-squares line for the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6) y 6 5 4 3 2 1 x Example 1, page 570

P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6) Solution Here, we have n = 5 and x1 = 1 x2 = 2 x3 = 3 x4 = 4 x5 = 5 y1 = 1 y2 = 3 y3 = 4 y4 = 3 y5 = 6 Substituting in the first equation we get Example 1, page 570

P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6) Solution Here, we have n = 5 and x1 = 1 x2 = 2 x3 = 3 x4 = 4 x5 = 5 y1 = 1 y2 = 3 y3 = 4 y4 = 3 y5 = 6 Substituting in the second equation we get Example 1, page 570

P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6) Solution Solving the simultaneous equations gives m = 1 and b = 0.4. Therefore, the required least-squares line is y = x + 0.4

P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6) Solution Below is the graph of the required least-squares line y = x + 0.4 y L 6 5 4 3 2 1 x Example 1, page 570

A market research study provided the following data based on the projected monthly sales x (in thousands) of an adventure movie DVD. Find the demand equation if the demand curve is the least-squares line for these data. The total monthly cost function associated with producing and distributing the DVD is given by C(x) = 4x + 25 where x denotes the number of discs (in thousands) produced and sold, and C(x) is in thousands of dollars. Determine the unit wholesale price that will maximize monthly profits. p 38 36 34.5 30 28.5 x 2.2 5.4 7.0 11.5 14.6 Applied Example 3, page 572

Solution The calculations required for obtaining the normal equations may be summarized as follows: Thus, the nominal equations are x p x2 xp 2.2 38.0 4.84 83.6 5.4 36.0 29.16 194.4 7.0 34.5 49.00 241.5 11.5 30.0 132.25 345.0 14.6 28.5 213.16 416.1 40.7 167.0 428.41 1280.6 Applied Example 3, page 572

Solution Solving the system of linear equations simultaneously, we find that Therefore, the required demand equation is given by Applied Example 3, page 572

Solution The total revenue function in this case is given by Since the total cost function is C(x) = 4x + 25 we see that the profit function is Applied Example 3, page 572

Solution To find the absolute maximum of P(x) over the closed interval [0, 49.37], we compute Since P(x) = 0 , we find that x ≈ as the only critical point of P. Finally, from the table we see that the optimal wholesale price is or $21.99 per disc. x 22.22 49.37 P(x) –25 374.78 – Applied Example 3, page 572

Constrained Maxima and Minima and the Method of Lagrange Multipliers
8.5 Constrained Maxima and Minima and the Method of Lagrange Multipliers

Constrained Maxima and Minima
In many practical optimization problems, we must maximize or minimize a function in which the independent variables are subjected to certain further constraints. We shall discuss a powerful method for determining relative extrema of a function f(x, y) whose independent variables x and y are required to satisfy one or more constraints of the form g(x, y) = 0.

Example Find the relative minimum of f(x, y) = 2x2 +y2 subject to the constraint g(x, y) = x + y – 1 = 0. Solution Solving the constraint equation for y explicitly in terms of x, we obtain y = – x + 1 Substituting this value of y into f(x, y) results in a function of x, Example 1, page 580

Example Find the relative minimum of f(x, y) = 2x2 +y2 subject to the constraint g(x, y) = x + y – 1 = 0. Solution We have h(x) = 3x2 – 2x + 1. The function h describes the curve lying on the graph of f on which the constrained relative minimum point (a, b) of f occurs: z f(x, y) = 2x2 + y2 h(x) = 3x2 – 2x + 1 (a, b, f(a, b)) y (a, b) g(x, y) = 0 Example 1, page 580 x

Example Find the relative minimum of f(x, y) = 2x2 +y2 subject to the constraint g(x, y) = x + y – 1 = 0. Solution To find this point (a, b), we determine the relative extrema of a function of one variable: Setting h′ = 0 gives x = as the sole critical point of h. Next, we find h″(x) = 6 and, in particular, h″( ) = 6 > 0. Therefore, by the second derivative test, the point gives rise to a relative minimum of h. Substitute x = into the constraint equation x +y – 1 = 0 to get y = Example 1, page 580

Example Find the relative minimum of f(x, y) = 2x2 +y2 subject to the constraint g(x, y) = x + y – 1 = 0. Solution Thus, the point ( ) gives rise to the required constrained relative minimum of f. Since , the required constrained relative minimum value of f is at the point It may be shown that is in fact a constrained absolute minimum value of f. Example 1, page 580

The Method of Lagrange Multipliers
To find the relative extrema of the function f(x, y) subject to the constraint g(x, y) = 0 (assuming that these extreme values exist), 1. Form an auxiliary function called the Lagrangian function (the variable λ is called the Lagrange multiplier). 2. Solve the system that consists of the equations Fx = Fy = Fλ = 0 for all values of x, y, and λ. 3. The solutions found in step 2 are candidates for the extrema of f.

Example Using the method of Lagrange multipliers, find the relative minimum of the function f(x, y) = 2x2 +y2 subject to the constraint x + y = 1. Solution Write the constraint equation x + y = 1 in the form g(x, y) = x + y – 1 = 0 Then, form the Lagrangian function Example 2, page 582

Example Using the method of Lagrange multipliers, find the relative minimum of the function f(x, y) = 2x2 +y2 subject to the constraint x + y = 1. Solution We have To find the critical point(s) of the function F, solve the system composed of the equations Solving the first and second equations for x and y in terms of λ, we obtain Which, upon substitution into the third equation yields Example 2, page 582

Example Using the method of Lagrange multipliers, find the relative minimum of the function f(x, y) = 2x2 +y2 subject to the constraint x + y = 1. Solution Substituting Therefore, and , and results in a constrained minimum of the function f. Example 2, page 582

Applied Example: Designing a Cruise-Ship Pool
The operators of the Viking Princess, a luxury cruise liner, are contemplating the addition of another swimming pool to the ship. The chief engineer has suggested that an area of the form of an ellipse located in the rear of the promenade deck would be suitable for this purpose. It has been determined that the shape of the ellipse may be described by the equation x2 + 4y2 = 3600 where x and y are measured in feet. Viking’s operators would like to know the dimensions of the rectangular pool with the largest possible area that would meet these requirements. Applied Example 5, page 582

Solution We want to maximize the area of the rectangle that will fit the ellipse: Letting the sides of the rectangle be 2x and 2y feet, we see that the area of the rectangle is A = 4xy. Furthermore, the point (x, y) must be constrained to lie on the ellipse so that it satisfies the equation x2 + 4y2 = 3600. y (x, y) x x2 + 4y2 = 3600 Applied Example 5, page 582

Solution Thus, the problem is equivalent to the problem of maximizing the function subject to the constraint g(x, y) = x2 + 4y2 – 3600 = 0 The Lagrangian function is To find the critical points of F, we solve the system of equations f(x, y) = 4xy Applied Example 5, page 582

Solution We have Solving the first equation for λ, we obtain Which, substituting into the second equation, yields Solving for x yields x = ± 2y. Applied Example 5, page 582

Solution We have Substituting x = ± 2y into the third equation, we have Which, upon solving for y yields The corresponding values of x are Since both x and y must be nonnegative, we have or approximately 42 ☓ 85 feet. Applied Example 5, page 582

8.6 Double Integrals

A Geometric Interpretation of the Double Integral
You may recall that we can do a Riemann sum to approximate the area under the graph of a function of one variable by adding the areas of the rectangles that form below the graph resulting from small increments of x (x) within a given interval [a, b]: y x y= f(x) x a b

Similarly, it is possible to obtain an approximation of the volume of the solid under the graph of a function of two variables. z y= f(x, y) c d y a b x

To find the volume of the solid under the surface, we can perform a Riemann sum of the volume Si of parallelepipeds with base Ri = x ☓ y and height f(xi, yi): z z = f(x, y) c d y a x y b x

To find the volume of the solid under the surface, we can perform a Riemann sum of the volume Si of parallelepipeds with base Ri = x ☓ y and height f(xi, yi): z z = f(x, y) Si c d y a x y b x

To find the volume of the solid under the surface, we can perform a Riemann sum of the volume Si of parallelepipeds with base Ri = x ☓ y and height f(xi, yi): z z = f(x, y) c d y a b x

The limit of the Riemann sum obtained when the number of rectangles m along the x-axis, and the number of subdivisions n along the y-axis tends to infinity is the value of the double integral of f(x, y) over the region R and is denoted by z z = f(x, y) c y ·n d y a x x ·m y b x

Theorem 1 Evaluating a Double Integral Over a Plane Region
Suppose g1(x) and g2(x) are continuous functions on [a, b] and the region R is defined by R = {(x, y)| g1(x)  y  g2(x); a  x  b}. Then, y y = g2(x) R y = g1(x) x a b

Theorem 1 Evaluating a Double Integral Over a Plane Region
Suppose h1(y) and h2(y) are continuous functions on [c, d] and the region R is defined by R = {(x, y)| h1(y)  x  h2(y); c  y  d}. Then, y d R x = h1(y) x = h2(y) c x

Examples Evaluate ∫R∫f(x, y)dA given that f(x, y) = x2 + y2 and R is the region bounded by the graphs of g1(x) = x and g2(x) = 2x for 0  x  2. Solution The region under consideration is: y g2(x) = 2x 4 3 2 1 g1(x) = x R x Example 2, page 593

Examples Evaluate ∫R∫f(x, y)dA given that f(x, y) = x2 + y2 and R is the region bounded by the graphs of g1(x) = x and g2(x) = 2x for 0  x  2. Solution Using Theorem 1, we find: Example 2, page 593

Examples Evaluate ∫R∫f(x, y)dA, where f(x, y) = xey and R is the plane region bounded by the graphs of y = x2 and y = x. Solution The region under consideration is: The points of intersection of the two curves are found by solving the equation x2 = x, giving x = 0 and x = 1. y 1 g2(x) = x R g1(x) = x2 x 1 Example 3, page 593

Integrating by parts on the right-hand side
Examples Evaluate ∫R∫f(x, y)dA, where f(x, y) = xey and R is the plane region bounded by the graphs of y = x2 and y = x. Solution Using Theorem 1, we find: Integrating by parts on the right-hand side Example 3, page 593

The Volume of a Solid Under a Surface
Let R be a region in the xy-plane and let f be continuous and nonnegative on R. Then, the volume of the solid under a surface bounded above by z = f(x, y) and below by R is given by

Example Find the volume of the solid bounded above by the plane z = f(x, y) = y and below by the plane region R defined by Solution The graph of the region R is: Observe that f(x, y) = y > 0 for (x, y) ∈ R. y 1 R x 1 Example 4, page 594

Example Find the volume of the solid bounded above by the plane z = f(x, y) = y and below by the plane region R defined by Solution Therefore, the required volume is given by Example 4, page 594

Example Find the volume of the solid bounded above by the plane z = f(x, y) = y and below by the plane region R defined by Solution z The graph of the solid in question is: z = f(x, y) = y y x Example 4, page 594

End of Chapter

8 Calculus of Several Variables Functions of Several Variables

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