Skip to Content
OpenStax Logo
Calculus Volume 3

4.8 Lagrange Multipliers

Calculus Volume 34.8 Lagrange Multipliers
Buy book
  1. Preface
  2. 1 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 1.1 Parametric Equations
    3. 1.2 Calculus of Parametric Curves
    4. 1.3 Polar Coordinates
    5. 1.4 Area and Arc Length in Polar Coordinates
    6. 1.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Vectors in Space
    1. Introduction
    2. 2.1 Vectors in the Plane
    3. 2.2 Vectors in Three Dimensions
    4. 2.3 The Dot Product
    5. 2.4 The Cross Product
    6. 2.5 Equations of Lines and Planes in Space
    7. 2.6 Quadric Surfaces
    8. 2.7 Cylindrical and Spherical Coordinates
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  4. 3 Vector-Valued Functions
    1. Introduction
    2. 3.1 Vector-Valued Functions and Space Curves
    3. 3.2 Calculus of Vector-Valued Functions
    4. 3.3 Arc Length and Curvature
    5. 3.4 Motion in Space
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  5. 4 Differentiation of Functions of Several Variables
    1. Introduction
    2. 4.1 Functions of Several Variables
    3. 4.2 Limits and Continuity
    4. 4.3 Partial Derivatives
    5. 4.4 Tangent Planes and Linear Approximations
    6. 4.5 The Chain Rule
    7. 4.6 Directional Derivatives and the Gradient
    8. 4.7 Maxima/Minima Problems
    9. 4.8 Lagrange Multipliers
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  6. 5 Multiple Integration
    1. Introduction
    2. 5.1 Double Integrals over Rectangular Regions
    3. 5.2 Double Integrals over General Regions
    4. 5.3 Double Integrals in Polar Coordinates
    5. 5.4 Triple Integrals
    6. 5.5 Triple Integrals in Cylindrical and Spherical Coordinates
    7. 5.6 Calculating Centers of Mass and Moments of Inertia
    8. 5.7 Change of Variables in Multiple Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Vector Calculus
    1. Introduction
    2. 6.1 Vector Fields
    3. 6.2 Line Integrals
    4. 6.3 Conservative Vector Fields
    5. 6.4 Green’s Theorem
    6. 6.5 Divergence and Curl
    7. 6.6 Surface Integrals
    8. 6.7 Stokes’ Theorem
    9. 6.8 The Divergence Theorem
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  8. 7 Second-Order Differential Equations
    1. Introduction
    2. 7.1 Second-Order Linear Equations
    3. 7.2 Nonhomogeneous Linear Equations
    4. 7.3 Applications
    5. 7.4 Series Solutions of Differential Equations
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 4.8.1. Use the method of Lagrange multipliers to solve optimization problems with one constraint.
  • 4.8.2. Use the method of Lagrange multipliers to solve optimization problems with two constraints.

Solving optimization problems for functions of two or more variables can be similar to solving such problems in single-variable calculus. However, techniques for dealing with multiple variables allow us to solve more varied optimization problems for which we need to deal with additional conditions or constraints. In this section, we examine one of the more common and useful methods for solving optimization problems with constraints.

Lagrange Multipliers

Example 4.41 was an applied situation involving maximizing a profit function, subject to certain constraints. In that example, the constraints involved a maximum number of golf balls that could be produced and sold in 11 month (x),(x), and a maximum number of advertising hours that could be purchased per month (y).(y). Suppose these were combined into a budgetary constraint, such as 20x+4y216,20x+4y216, that took into account the cost of producing the golf balls and the number of advertising hours purchased per month. The goal is, still, to maximize profit, but now there is a different type of constraint on the values of xx and y.y. This constraint, when combined with the profit function f(x,y)=48x+96yx22xy9y2,f(x,y)=48x+96yx22xy9y2, is an example of an optimization problem, and the function f(x,y)f(x,y) is called the objective function. A graph of various level curves of the function f(x,y)f(x,y) follows.

A series of rotated ellipses that become increasingly large. The smallest one is marked f(x, y) = 400, and the biggest one is marked f(x, y) = 150.
Figure 4.59 Graph of level curves of the function f(x,y)=48x+96yx22xy9y2f(x,y)=48x+96yx22xy9y2 corresponding to c=150,250,350,and400.c=150,250,350,and400.

In Figure 4.59, the value cc represents different profit levels (i.e., values of the function f).f). As the value of cc increases, the curve shifts to the right. Since our goal is to maximize profit, we want to choose a curve as far to the right as possible. If there was no restriction on the number of golf balls the company could produce, or the number of units of advertising available, then we could produce as many golf balls as we want, and advertise as much as we want, and there would be not be a maximum profit for the company. Unfortunately, we have a budgetary constraint that is modeled by the inequality 20x+4y216.20x+4y216. To see how this constraint interacts with the profit function, Figure 4.60 shows the graph of the line 20x+4y=21620x+4y=216 superimposed on the previous graph.

A series of rotated ellipses that become increasingly large. On the smallest ellipse, which is red, there is a tangent line marked with equation 20x + 4y = 216 that appears to touch the ellipse near (10, 4).
Figure 4.60 Graph of level curves of the function f(x,y)=48x+96yx22xy9y2f(x,y)=48x+96yx22xy9y2 corresponding to c=150,250,350,and395.c=150,250,350,and395. The red graph is the constraint function.

As mentioned previously, the maximum profit occurs when the level curve is as far to the right as possible. However, the level of production corresponding to this maximum profit must also satisfy the budgetary constraint, so the point at which this profit occurs must also lie on (or to the left of) the red line in Figure 4.60. Inspection of this graph reveals that this point exists where the line is tangent to the level curve of f.f. Trial and error reveals that this profit level seems to be around 395,395, when xx and yy are both just less than 5.5. We return to the solution of this problem later in this section. From a theoretical standpoint, at the point where the profit curve is tangent to the constraint line, the gradient of both of the functions evaluated at that point must point in the same (or opposite) direction. Recall that the gradient of a function of more than one variable is a vector. If two vectors point in the same (or opposite) directions, then one must be a constant multiple of the other. This idea is the basis of the method of Lagrange multipliers.

Theorem 4.20

Method of Lagrange Multipliers: One Constraint

Let ff and gg be functions of two variables with continuous partial derivatives at every point of some open set containing the smooth curve g(x,y)=0.g(x,y)=0. Suppose that f,f, when restricted to points on the curve g(x,y)=0,g(x,y)=0, has a local extremum at the point (x0,y0)(x0,y0) and that g(x0,y0)0.g(x0,y0)0. Then there is a number λλ called a Lagrange multiplier, for which

f(x0,y0)=λg(x0,y0).f(x0,y0)=λg(x0,y0).

Proof

Assume that a constrained extremum occurs at the point (x0,y0).(x0,y0). Furthermore, we assume that the equation g(x,y)=0g(x,y)=0 can be smoothly parameterized as

x=x(s)andy=y(s)x=x(s)andy=y(s)

where s is an arc length parameter with reference point (x0,y0)(x0,y0) at s=0.s=0. Therefore, the quantity z=f(x(s),y(s))z=f(x(s),y(s)) has a relative maximum or relative minimum at s=0,s=0, and this implies that dzds=0dzds=0 at that point. From the chain rule,

dzds=fx·xs+fy·ys=( fxî+ fy)·(xsî·ys)=0,dzds=fx·xs+fy·ys=( fxî+ fy)·(xsî·ys)=0,

where the derivatives are all evaluated at s=0.s=0. However, the first factor in the dot product is the gradient of f,f, and the second factor is the unit tangent vector T(0)T(0) to the constraint curve. Since the point (x0,y0)(x0,y0) corresponds to s=0,s=0, it follows from this equation that

f(x0,y0)·T(0)=0,f(x0,y0)·T(0)=0,

which implies that the gradient is either 00 or is normal to the constraint curve at a constrained relative extremum. However, the constraint curve g(x,y)=0g(x,y)=0 is a level curve for the function g(x,y)g(x,y) so that if g(x0,y0)0g(x0,y0)0 then g(x0,y0)g(x0,y0) is normal to this curve at (x0,y0)(x0,y0) It follows, then, that there is some scalar λλ such that

f(x0,y0)=λg(x0,y0)f(x0,y0)=λg(x0,y0)

To apply Method of Lagrange Multipliers: One Constraint to an optimization problem similar to that for the golf ball manufacturer, we need a problem-solving strategy.

Problem-Solving Strategy: Steps for Using Lagrange Multipliers
  1. Determine the objective function f(x,y)f(x,y) and the constraint function g(x,y).g(x,y). Does the optimization problem involve maximizing or minimizing the objective function?
  2. Set up a system of equations using the following template:
    f(x0,y0)=λg(x0,y0)g(x0,y0)=0.f(x0,y0)=λg(x0,y0)g(x0,y0)=0.
  3. Solve for x0x0 and y0.y0.
  4. The largest of the values of ff at the solutions found in step 33 maximizes f;f; the smallest of those values minimizes f.f.

Example 4.42

Using Lagrange Multipliers

Use the method of Lagrange multipliers to find the minimum value of f(x,y)=x2+4y22x+8yf(x,y)=x2+4y22x+8y subject to the constraint x+2y=7.x+2y=7.

Solution

Let’s follow the problem-solving strategy:

  1. The optimization function is f(x,y)=x2+4y22x+8y.f(x,y)=x2+4y22x+8y. To determine the constraint function, we must first subtract 77 from both sides of the constraint. This gives x+2y7=0.x+2y7=0. The constraint function is equal to the left-hand side, so g(x,y)=x+2y7.g(x,y)=x+2y7. The problem asks us to solve for the minimum value of f,f, subject to the constraint (see the following graph).
    Two rotated ellipses, one within the other. On the largest ellipse, which is marked f(x, y) = 26, there is a tangent line marked with equation x + 2y = 7 that appears to touch the ellipse near (5, 1).
    Figure 4.61 Graph of level curves of the function f(x,y)=x2+4y22x+8yf(x,y)=x2+4y22x+8y corresponding to c=10c=10 and 26.26. The red graph is the constraint function.
  2. We then must calculate the gradients of both f and g:
    f(x,y)=(2x2)i+(8y+8)jg(x,y)=i+2j.f(x,y)=(2x2)i+(8y+8)jg(x,y)=i+2j.

    The equation f(x0,y0)=λg(x0,y0)f(x0,y0)=λg(x0,y0) becomes
    (2x02)i+(8y0+8)j=λ(i+2j),(2x02)i+(8y0+8)j=λ(i+2j),

    which can be rewritten as
    (2x02)i+(8y0+8)j=λi+j.(2x02)i+(8y0+8)j=λi+j.

    Next, we set the coefficients of iandjiandj equal to each other:
    2x02=λ8y0+8=2λ.2x02=λ8y0+8=2λ.

    The equation g(x0,y0)=0g(x0,y0)=0 becomes x0+2y07=0.x0+2y07=0. Therefore, the system of equations that needs to be solved is
    2x02=λ8y0+8=2λx0+2y07=0.2x02=λ8y0+8=2λx0+2y07=0.
  3. This is a linear system of three equations in three variables. We start by solving the second equation for λλ and substituting it into the first equation. This gives λ=4y0+4,λ=4y0+4, so substituting this into the first equation gives
    2x02=4y0+4.2x02=4y0+4.

    Solving this equation for x0x0 gives x0=2y0+3.x0=2y0+3. We then substitute this into the third equation:
    (2y0+3)+2y07=04y04=0y0=1.(2y0+3)+2y07=04y04=0y0=1.

    Since x0=2y0+3,x0=2y0+3, this gives x0=5.x0=5.
  4. Next, we substitute (5,1)(5,1) into f(x,y)=x2+4y22x+8y,f(x,y)=x2+4y22x+8y, gives f(5,1)=52+4(1)22(5)+8(1)=27.f(5,1)=52+4(1)22(5)+8(1)=27. To ensure this corresponds to a minimum value on the constraint function, let’s try some other values, such as the intercepts of g(x,y)=0,g(x,y)=0, Which are (7,0)(7,0) and (0,3.5).(0,3.5). We get f(7,0)=35f(7,0)=35 and f(0,3.5)=77,f(0,3.5)=77, so it appears ff has a minimum at (5,1).(5,1).

Checkpoint 4.37

Use the method of Lagrange multipliers to find the maximum value of f(x,y)=9x2+36xy4y218x8yf(x,y)=9x2+36xy4y218x8y subject to the constraint 3x+4y=32.3x+4y=32.

Let’s now return to the problem posed at the beginning of the section.

Example 4.43

Golf Balls and Lagrange Multipliers

The golf ball manufacturer, Pro-T, has developed a profit model that depends on the number xx of golf balls sold per month (measured in thousands), and the number of hours per month of advertising y, according to the function

z=f(x,y)=48x+96yx22xy9y2,z=f(x,y)=48x+96yx22xy9y2,

where zz is measured in thousands of dollars. The budgetary constraint function relating the cost of the production of thousands golf balls and advertising units is given by 20x+4y=216.20x+4y=216. Find the values of xx and yy that maximize profit, and find the maximum profit.

Solution

Again, we follow the problem-solving strategy:

  1. The optimization function is f(x,y)=48x+96yx22xy9y2.f(x,y)=48x+96yx22xy9y2. To determine the constraint function, we first subtract 216 from both sides of the constraint, then divide both sides by 4,4, which gives 5x+y54=0.5x+y54=0. The constraint function is equal to the left-hand side, so g(x,y)=5x+y54.g(x,y)=5x+y54. The problem asks us to solve for the maximum value of f,f, subject to this constraint.
  2. So, we calculate the gradients of both fandfand g:g:
    f(x,y)=(482x2y)i+(962x18y)jg(x,y)=5i+j.f(x,y)=(482x2y)i+(962x18y)jg(x,y)=5i+j.

    The equation f(x0,y0)=λg(x0,y0)f(x0,y0)=λg(x0,y0) becomes
    (482x02y0)i+(962x018y0)j=λ(5i+j),(482x02y0)i+(962x018y0)j=λ(5i+j),

    which can be rewritten as
    (482x02y0)i+(962x018y0)j=λ5i+λj.(482x02y0)i+(962x018y0)j=λ5i+λj.

    We then set the coefficients of iandjiandj equal to each other:
    482x02y0=5λ962x018y0=λ.482x02y0=5λ962x018y0=λ.

    The equation g(x0,y0)=0g(x0,y0)=0 becomes 5x0+y054=0.5x0+y054=0. Therefore, the system of equations that needs to be solved is
    482x02y0=5λ962x018y0=λ5x0+y054=0.482x02y0=5λ962x018y0=λ5x0+y054=0.
  3. We use the left-hand side of the second equation to replace λλ in the first equation:
    482x02y0=5(962x018y0)482x02y0=48010x090y08x0=43288y0x0=5411y0.482x02y0=5(962x018y0)482x02y0=48010x090y08x0=43288y0x0=5411y0.

    Then we substitute this into the third equation:
    5(5411y0)+y054=027055y0+y0=021654y0=0y0=4.5(5411y0)+y054=027055y0+y0=021654y0=0y0=4.

    Since x0=5411y0,x0=5411y0, this gives x0=10.x0=10.
  4. We then substitute (10,4)(10,4) into f(x,y)=48x+96yx22xy9y2,f(x,y)=48x+96yx22xy9y2, which gives
    f(10,4)=48(10)+96(4)(10)22(10)(4)9(4)2=480+38410080144=540.f(10,4)=48(10)+96(4)(10)22(10)(4)9(4)2=480+38410080144=540.

    Therefore the maximum profit that can be attained, subject to budgetary constraints, is $540,000$540,000 with a production level of 10,00010,000 golf balls and 44 hours of advertising bought per month. Let’s check to make sure this truly is a maximum. The endpoints of the line that defines the constraint are (10.8,0)(10.8,0) and (0,54)(0,54) Let’s evaluate ff at both of these points:
    f(10.8,0)=48(10.8)+96(0)10.822(10.8)(0)9(02)=401.76f(0,54)=48(0)+96(54)022(0)(54)9(542)=−21,060.f(10.8,0)=48(10.8)+96(0)10.822(10.8)(0)9(02)=401.76f(0,54)=48(0)+96(54)022(0)(54)9(542)=−21,060.

    The second value represents a loss, since no golf balls are produced. Neither of these values exceed 540,540, so it seems that our extremum is a maximum value of f.f.

Checkpoint 4.38

A company has determined that its production level is given by the Cobb-Douglas function f(x,y)=2.5x0.45y0.55f(x,y)=2.5x0.45y0.55 where x represents the total number of labor hours in 11 year and y represents the total capital input for the company. Suppose 11 unit of labor costs $40$40 and 11 unit of capital costs $50.$50. Use the method of Lagrange multipliers to find the maximum value of f(x,y)=2.5x0.45y0.55f(x,y)=2.5x0.45y0.55 subject to a budgetary constraint of $500,000$500,000 per year.

In the case of an optimization function with three variables and a single constraint function, it is possible to use the method of Lagrange multipliers to solve an optimization problem as well. An example of an optimization function with three variables could be the Cobb-Douglas function in the previous example: f(x,y,z)=x0.2y0.4z0.4,f(x,y,z)=x0.2y0.4z0.4, where xx represents the cost of labor, yy represents capital input, and zz represents the cost of advertising. The method is the same as for the method with a function of two variables; the equations to be solved are

f(x,y,z)=λg(x,y,z)g(x,y,z)=0.f(x,y,z)=λg(x,y,z)g(x,y,z)=0.

Example 4.44

Lagrange Multipliers with a Three-Variable Optimization Function

Maximize the function f(x,y,z)=x2+y2+z2f(x,y,z)=x2+y2+z2 subject to the constraint x+y+z=1.x+y+z=1.

Solution

  1. The optimization function is f(x,y,z)=x2+y2+z2.f(x,y,z)=x2+y2+z2. To determine the constraint function, we subtract 11 from each side of the constraint: x+y+z1=0x+y+z1=0 which gives the constraint function as g(x,y,z)=x+y+z1.g(x,y,z)=x+y+z1.
  2. Next, we calculate f(x,y,z)f(x,y,z) and g(x,y,z):g(x,y,z):
    f(x,y,z)=2x,2y,2zg(x,y,z)=1,1,1.f(x,y,z)=2x,2y,2zg(x,y,z)=1,1,1.

    This leads to the equations
    2x0,2y0,2z0=λ1,1,1x0+y0+z01=02x0,2y0,2z0=λ1,1,1x0+y0+z01=0

    which can be rewritten in the following form:
    2x0=λ2y0=λ2z0=λx0+y0+z01=0.2x0=λ2y0=λ2z0=λx0+y0+z01=0.
  3. Since each of the first three equations has λλ on the right-hand side, we know that 2x0=2y0=2z02x0=2y0=2z0 and all three variables are equal to each other. Substituting y0=x0y0=x0 and z0=x0z0=x0 into the last equation yields 3x01=0,3x01=0, so x0=13x0=13 and y0=13y0=13 and z0=13z0=13 which corresponds to a critical point on the constraint curve.
  4. Then, we evaluate f at the point (13,13,13):(13,13,13):
    f(13,13,13)=(13)2+(13)2+(13)2=39=13.f(13,13,13)=(13)2+(13)2+(13)2=39=13.

    Therefore, an extremum of the function is 13.13. To verify it is a minimum, choose other points that satisfy the constraint and calculate ff at that point. For example,
    f(1,0,0)=12+02+02=1f(0,−2,3)=02+(−2)2+32=13.f(1,0,0)=12+02+02=1f(0,−2,3)=02+(−2)2+32=13.

    Both of these values are greater than 13,13, leading us to believe the extremum is a minimum.

Checkpoint 4.39

Use the method of Lagrange multipliers to find the minimum value of the function

f(x,y,z)=x+y+zf(x,y,z)=x+y+z

subject to the constraint x2+y2+z2=1.x2+y2+z2=1.

Problems with Two Constraints

The method of Lagrange multipliers can be applied to problems with more than one constraint. In this case the optimization function, ww is a function of three variables:

w=f(x,y,z)w=f(x,y,z)

and it is subject to two constraints:

g(x,y,z)=0andh(x,y,z)=0.g(x,y,z)=0andh(x,y,z)=0.

There are two Lagrange multipliers, λ1λ1 and λ2,λ2, and the system of equations becomes

f(x0,y0,z0)=λ1g(x0,y0,z0)+λ2h(x0,y0,z0)g(x0,y0,z0)=0h(x0,y0,z0)=0.f(x0,y0,z0)=λ1g(x0,y0,z0)+λ2h(x0,y0,z0)g(x0,y0,z0)=0h(x0,y0,z0)=0.

Example 4.45

Lagrange Multipliers with Two Constraints

Find the maximum and minimum values of the function

f(x,y,z)=x2+y2+z2f(x,y,z)=x2+y2+z2

subject to the constraints z2=x2+y2z2=x2+y2 and x+yz+1=0.x+yz+1=0.

Solution

Let’s follow the problem-solving strategy:

  1. The optimization function is f(x,y,z)=x2+y2+z2.f(x,y,z)=x2+y2+z2. To determine the constraint functions, we first subtract z2z2 from both sides of the first constraint, which gives x2+y2z2=0,x2+y2z2=0, so g(x,y,z)=x2+y2z2.g(x,y,z)=x2+y2z2. The second constraint function is h(x,y,z)=x+yz+1.h(x,y,z)=x+yz+1.
  2. We then calculate the gradients of f,g,andh:f,g,andh:
    f(x,y,z)=2xi+2yj+2zkg(x,y,z)=2xi+2yj2zkh(x,y,z)=i+jk.f(x,y,z)=2xi+2yj+2zkg(x,y,z)=2xi+2yj2zkh(x,y,z)=i+jk.

    The equation f(x0,y0,z0)=λ1g(x0,y0,z0)+λ2h(x0,y0,z0)f(x0,y0,z0)=λ1g(x0,y0,z0)+λ2h(x0,y0,z0) becomes
    2x0i+2y0j+2z0k=λ1(2x0i+2y0j2z0k)+λ2(i+jk),2x0i+2y0j+2z0k=λ1(2x0i+2y0j2z0k)+λ2(i+jk),

    which can be rewritten as
    2x0i+2y0j+2z0k=(2λ1x0+λ2)i+(2λ1y0+λ2)j(2λ1z0+λ2)k.2x0i+2y0j+2z0k=(2λ1x0+λ2)i+(2λ1y0+λ2)j(2λ1z0+λ2)k.

    Next, we set the coefficients of i,j, andki,j, andk equal to each other:
    2x0=2λ1x0+λ22y0=2λ1y0+λ22z0=−2λ1z0λ2.2x0=2λ1x0+λ22y0=2λ1y0+λ22z0=−2λ1z0λ2.

    The two equations that arise from the constraints are z02=x02+y02z02=x02+y02 and x0+y0z0+1=0.x0+y0z0+1=0. Combining these equations with the previous three equations gives
    2x0=2λ1x0+λ22y0=2λ1y0+λ22z0=−2λ1z0λ2z02=x02+y02x0+y0z0+1=0.2x0=2λ1x0+λ22y0=2λ1y0+λ22z0=−2λ1z0λ2z02=x02+y02x0+y0z0+1=0.
  3. The first three equations contain the variable λ2.λ2. Solving the third equation for λ2λ2 and replacing into the first and second equations reduces the number of equations to four:
    2x0=2λ1x02λ1z02z02y0=2λ1y02λ1z02z0z02=x02+y02x0+y0z0+1=0.2x0=2λ1x02λ1z02z02y0=2λ1y02λ1z02z0z02=x02+y02x0+y0z0+1=0.

    Next, we solve the first and second equation for λ1.λ1. The first equation gives λ1=x0+z0x0z0,λ1=x0+z0x0z0, the second equation gives λ1=y0+z0y0z0.λ1=y0+z0y0z0. We set the right-hand side of each equation equal to each other and cross-multiply:
    x0+z0x0z0=y0+z0y0z0(x0+z0)(y0z0)=(x0z0)(y0+z0)x0y0x0z0+y0z0z02=x0y0+x0z0y0z0z022y0z02x0z0=02z0(y0x0)=0..x0+z0x0z0=y0+z0y0z0(x0+z0)(y0z0)=(x0z0)(y0+z0)x0y0x0z0+y0z0z02=x0y0+x0z0y0z0z022y0z02x0z0=02z0(y0x0)=0..

    Therefore, either z0=0z0=0 or y0=x0.y0=x0. If z0=0,z0=0, then the first constraint becomes 0=x02+y02.0=x02+y02. The only real solution to this equation is x0=0x0=0 and y0=0,y0=0, which gives the ordered triple (0,0,0).(0,0,0). This point does not satisfy the second constraint, so it is not a solution.
    Next, we consider y0=x0,y0=x0, which reduces the number of equations to three:
    y0=x0z02=x02+y02x0+y0z0+1=0.y0=x0z02=x02+y02x0+y0z0+1=0.

    We substitute the first equation into the second and third equations:
    z02=x02+x02x0+x0z0+1=0.z02=x02+x02x0+x0z0+1=0.

    Then, we solve the second equation for z0,z0, which gives z0=2x0+1.z0=2x0+1. We then substitute this into the first equation,
    z02=2x02(2x0+1)2=2x024x02+4x0+1=2x022x02+4x0+1=0,z02=2x02(2x0+1)2=2x024x02+4x0+1=2x022x02+4x0+1=0,

    and use the quadratic formula to solve for x0:x0:
    x0=−4±424(2)(1)2(2)=−4±84=−4±224=−1±22.x0=−4±424(2)(1)2(2)=−4±84=−4±224=−1±22.

    Recall y0=x0,y0=x0, so this solves for y0y0 as well. Then, z0=2x0+1,z0=2x0+1, so
    z0=2x0+1=2(−1±22)+1=−2+1±2=−1±2.z0=2x0+1=2(−1±22)+1=−2+1±2=−1±2.

    Therefore, there are two ordered triplet solutions:
    (−1+22,−1+22,−1+2)and(−122,−122,−12).(−1+22,−1+22,−1+2)and(−122,−122,−12).
  4. We substitute (−1+22,−1+22,−1+2)(−1+22,−1+22,−1+2) into f(x,y,z)=x2+y2+z2,f(x,y,z)=x2+y2+z2, which gives
    f(−1+22,−1+22,−1+2)=(−1+22)2+(−1+22)2+(−1+2)2=(12+12)+(12+12)+(122+2)=642.f(−1+22,−1+22,−1+2)=(−1+22)2+(−1+22)2+(−1+2)2=(12+12)+(12+12)+(122+2)=642.

    Then, we substitute (−122,−122,−12)(−122,−122,−12) into f(x,y,z)=x2+y2+z2,f(x,y,z)=x2+y2+z2, which gives
    f(−122,−122,−12)=(−122)2+(−122)2+(−12)2=(1+2+12)+(1+2+12)+(1+22+2)=6+42.f(−122,−122,−12)=(−122)2+(−122)2+(−12)2=(1+2+12)+(1+2+12)+(1+22+2)=6+42.

    6+426+42 is the maximum value and 642642 is the minimum value of f(x,y,z),f(x,y,z), subject to the given constraints.
Checkpoint 4.40

Use the method of Lagrange multipliers to find the minimum value of the function

f(x,y,z)=x2+y2+z2f(x,y,z)=x2+y2+z2

subject to the constraints 2x+y+2z=92x+y+2z=9 and 5x+5y+7z=29.5x+5y+7z=29.

Section 4.8 Exercises

For the following exercises, use the method of Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints.

358.

f(x,y)=x2y;x2+2y2=6f(x,y)=x2y;x2+2y2=6

359.

f(x,y,z)=xyz,x2+2y2+3z2=6f(x,y,z)=xyz,x2+2y2+3z2=6

360.

f(x,y)=xy;4x2+8y2=16f(x,y)=xy;4x2+8y2=16

361.

f(x,y)=4x3+y2;2x2+y2=1f(x,y)=4x3+y2;2x2+y2=1

362.

f(x,y,z)=x2+y2+z2,x4+y4+z4=1f(x,y,z)=x2+y2+z2,x4+y4+z4=1

363.

f(x,y,z)=yz+xy,xy=1,y2+z2=1f(x,y,z)=yz+xy,xy=1,y2+z2=1

364.

f(x,y)=x2+y2,(x1)2+4y2=4f(x,y)=x2+y2,(x1)2+4y2=4

365.

f(x,y)=4xy,x29+y216=1f(x,y)=4xy,x29+y216=1

366.

f(x,y,z)=x+y+z,1x+1y+1z=1f(x,y,z)=x+y+z,1x+1y+1z=1

367.

f(x,y,z)=x+3yz,x2+y2+z2=4f(x,y,z)=x+3yz,x2+y2+z2=4

368.

f(x,y,z)=x2+y2+z2,xyz=4f(x,y,z)=x2+y2+z2,xyz=4

369.

Minimize f(x,y)=x2+y2f(x,y)=x2+y2 on the hyperbola xy=1.xy=1.

370.

Minimize f(x,y)=xyf(x,y)=xy on the ellipse b2x2+a2y2=a2b2.b2x2+a2y2=a2b2.

371.

Maximize f(x,y,z)=2x+3y+5zf(x,y,z)=2x+3y+5z on the sphere x2+y2+z2=19.x2+y2+z2=19.

372.

Maximize f(x,y)=x2y2;x>0,y>0;g(x,y)=yx2=0f(x,y)=x2y2;x>0,y>0;g(x,y)=yx2=0

373.

The curve x3y3=1x3y3=1 is asymptotic to the line y=x.y=x. Find the point(s) on the curve x3y3=1x3y3=1 farthest from the line y=x.y=x.

374.

Maximize U(x,y)=8x4/5y1/5;4x+2y=12U(x,y)=8x4/5y1/5;4x+2y=12

375.

Minimize f(x,y)=x2+y2,x+2y5=0.f(x,y)=x2+y2,x+2y5=0.

376.

Maximize f(x,y)=6x2y2,x+y2=0.f(x,y)=6x2y2,x+y2=0.

377.

Minimize f(x,y,z)=x2+y2+z2,x+y+z=1.f(x,y,z)=x2+y2+z2,x+y+z=1.

378.

Minimize f(x,y)=x2y2f(x,y)=x2y2 subject to the constraint x2y+6=0.x2y+6=0.

379.

Minimize f(x,y,z)=x2+y2+z2f(x,y,z)=x2+y2+z2 when x+y+z=9x+y+z=9 and x+2y+3z=20.x+2y+3z=20.

For the next group of exercises, use the method of Lagrange multipliers to solve the following applied problems.

380.

A pentagon is formed by placing an isosceles triangle on a rectangle, as shown in the diagram. If the perimeter of the pentagon is 1010 in., find the lengths of the sides of the pentagon that will maximize the area of the pentagon.

A rectangle with an isosceles triangle on top. The side of the isosceles triangle with the two equal angles of size θ overlaps the top length of the rectangle.
381.

A rectangular box without a top (a topless box) is to be made from 1212 ft2 of cardboard. Find the maximum volume of such a box.

382.

Find the minimum and maximum distances between the ellipse x2+xy+2y2=1x2+xy+2y2=1 and the origin.

383.

Find the point on the surface x22xy+y2x+y=0x22xy+y2x+y=0 closest to the point (1,2,−3).(1,2,−3).

384.

Show that, of all the triangles inscribed in a circle of radius RR (see diagram), the equilateral triangle has the largest perimeter.

A circle with an equilateral triangle drawn inside of it such that each vertex of the triangle touches the circle.
385.

Find the minimum distance from point (0,1)(0,1) to the parabola x2=4y.x2=4y.

386.

Find the minimum distance from the parabola y=x2y=x2 to point (0,3).(0,3).

387.

Find the minimum distance from the plane x+y+z=1x+y+z=1 to point (2,1,1).(2,1,1).

388.

A large container in the shape of a rectangular solid must have a volume of 480480 m3. The bottom of the container costs $5/m2 to construct whereas the top and sides cost $3/m2 to construct. Use Lagrange multipliers to find the dimensions of the container of this size that has the minimum cost.

389.

Find the point on the line y=2x+3y=2x+3 that is closest to point (4,2).(4,2).

390.

Find the point on the plane 4x+3y+z=24x+3y+z=2 that is closest to the point (1,−1,1).(1,−1,1).

391.

Find the maximum value of f(x,y)=sinxsiny,f(x,y)=sinxsiny, where xandyxandy denote the acute angles of a right triangle. Draw the contours of the function using a CAS.

392.

A rectangular solid is contained within a tetrahedron with vertices at

(1,0,0),(0,1,0),(0,0,1),(1,0,0),(0,1,0),(0,0,1), and the origin. The base of the box has dimensions x,y,x,y, and the height of the box is z.z. If the sum of x,y,andzx,y,andz is 1.0, find the dimensions that maximizes the volume of the rectangular solid.

393.

[T] By investing x units of labor and y units of capital, a watch manufacturer can produce P(x,y)=50x0.4y0.6P(x,y)=50x0.4y0.6 watches. Find the maximum number of watches that can be produced on a budget of $20,000$20,000 if labor costs $100/unit and capital costs $200/unit. Use a CAS to sketch a contour plot of the function.

Citation/Attribution

Want to cite, share, or modify this book? This book is Creative Commons Attribution-NonCommercial-ShareAlike License 4.0 and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/calculus-volume-3/pages/1-introduction
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/calculus-volume-3/pages/1-introduction
Citation information

© Mar 30, 2016 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License 4.0 license. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.