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Mathematics LibreTexts

Exercises for Lagrange Multipliers

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In exercises 1-15, use the method of Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint.

1) Objective function: f(x,y)=4xy Constraint: x29+y216=1

Answer
Subject to the given constraint, the function f has a relative minimum of 24 at both (322,22) and (322,22) and a relative maximum of 24 at both (322,22) and (322,22)

2) Objective function: f(x,y)=x2y Constraint: x2+2y2=6

3) Objective function: f(x,y)=x2+y2+2x2y+1 Constraint: g(x,y)=x2+y2=2

Answer
Subject to the given constraint, f has a relative minimum of 1 at (1,1) and a relative maximum of 7 at (1,1).

4) Objective function: f(x,y)=xy Constraint: 4x2+8y2=16

5) Objective function: f(x,y)=x2+y2 Constraint: xy=1

Answer
f has a relative minimum of 2 at both (1,1) and (1,1), subject to the given constraint.

6) Objective function: f(x,y)=x2y2 Constraint: x2y+6=0

7) Objective function: f(x,y)=x2+y2 Constraint: x+2y5=0

Answer
Subject to the given constraint, f has a relative minimum of f(1,2)=5 at the point (1,2).

8) Objective function: f(x,y)=x2+y2 Constraint: (x1)2+4y2=4

9) Objective function: f(x,y)=4x3+y2 Constraint: 2x2+y2=1

Answer
Subject to the given constraint, the function f has a relative minimum of 2 at (22,0),
a relative minimum of 2527 at both points (13,73) and (13,73),
a relative maximum of 2 at (22,0), and a relative maximum of 1 at both points (0,1) and (0,1).
Solution:
Let g(x,y)=2x2+y2 be the constraint function. Then:

f(x,y)=12x2ˆi+2yˆj and g(x,y)=4xˆi+2yˆj

Using the Lagrange Multiplier equation, f(x,y)=λg(x,y),
we have: 12x2ˆi+2yˆj=4xλˆi+2yλˆj
giving us the system of equations: 12x2=4xλ,2y=2yλ,and the constraint2x2+y2=1
Rewriting the first two equations as zero-products (by moving to one-side and factoring), we get:
4x(3xλ)=0and2y(1λ)=0x=0orλ=3xy=0orλ=1
Now we consider the combinations of these solutions to the above two equations and plug each of these into the constraint equation to solve for the corresponding Lagrange points.

The combination x=0 and y=0 produces a contradiction when placed in the constraint equation, since this point is not on the ellipse.

Taking the combination x=0 and λ=1, we put 0 in for x in the constraint and solve for y, obtaining: y=±1. This gives us two Lagrange points: (0,1) and (0,1).

Taking the combination λ=3x and y=0, we put 0 in for y in the constraint and solve for x, obtaining: x=±22. This gives us two Lagrange points: (22,0) and (22,0).

Taking the combination λ=3x and λ=1, we substitute 1 into the first equation for λ, giving us 1=3x so x=13. Plugging this value in for x in the constraint equation and solving for y, we obtain y=±73 which gives us the two Lagrange points: (13,73) and (13,73).

Evaluating the function f at these Lagrange points, we find: f(0,1)=1f(0,1)=1f(22,0)=4(2)38=2f(22,0)=4(2)38=2f(13,73)=2527f(13,73)=2527

Comparing these values with where the corresponding Lagrange points lie on the constraint curve, we conclude the results stated in the answer above.

10) Objective function: f(x,y)=2x2+y2 Constraint: g(x,y)=x2+y2=1

11) Objective function: f(x,y,z)=x+3yz Constraint: x2+y2+z2=4

Answer
Subject to the given constraint,f has a relative minimum of 211 at the point (21111,61111,21111) and a relative maximum of 211 at the point (21111,61111,21111).

12) Objective function: f(x,y,z)=x+y+z Constraint: 1x+1y+1z=1

13) Objective function: f(x,y,z)=xyz Constraint: x2+2y2+3z2=6

Answer
Subject to the given constraint, f has a relative minimum of 233 at (2,1,63),(2,1,63),(2,1,63), and (2,1,63) and a relative maximum of 233 at (2,1,63),(2,1,63),(2,1,63), and (2,1,63).

14) Objective function: f(x,y,z)=x2+y2+z2 Constraint: x4+y4+z4=1

15) Objective function: f(x,y,z)=x2+y2+z2 Constraint: xyz=4

Answer
Subject to the given constraint, f has a relative minimum of 632 at the points (34,34,34), (34,34,34), (34,34,34), and (34,34,34).
To see a 3D visualization of this problem, see: CalcPlot3D for Problem 15.

In exercises 16-21, use the method of Lagrange multipliers to find the requested extremum of the given function subject to the given constraint.

16) Maximize f(x,y)=6x2y2 subject to the constraint, x+y2=0.

17) Maximize f(x,y)=x2y2 subject to the constraints, g(x,y)=yx2=0,x>0,y>0.

Answer
Subject to the given constraints, f has a relative maximum of f(22,12)=14 at the point (22,12). If x>0 were not a constraint, there would have been two other Lagrange points with relative extrema of f subject to the other two constraints. These would have been (0,0) and (22,12).

To verify that f really has a relative maximum at the point (22,12), we would need to check the value of f on either side of this point on the constraint curve, yx2=0.

If x=0.5 which is less than 22, y would be y=(0.5)2=0.25.
If x=1 which is greater than 22, y would be y=(1)2=1.

Then we compare the value of f at the Lagrange point, (22,12), with the values of f at these other points on the constraint.
We have f(0.5,0.25)=(0.5)2(0.25)2=0.250.0625=0.1875<14 and f(1,1)=(1)2(1)2=0<14.

Therefore, we can conclude that f indeed has a relative maximum of 14 at the point (22,12).

18) Maximize U(x,y)=8x4/5y1/5 subject to the constraint, 4x+2y=12.

19) Minimize f(x,y,z)=x2+y2+z2 subject to the constraint, x+y+z=1.

Answer
Subject to the given constraint, f has a relative minimum of f(13,13,13)=13 at the point (13,13,13).

20) Minimize f(x,y)=xy on the ellipse x2a2+y2b2=1.

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

Answer
Subject to the given constraint, f has a relative maximum of 192 at the point (2,322,522).
Note that, subject to this constraint, f also has a relative minimum of 192 at the point (2,322,522).
To see a 3D visualization of this problem, see: CalcPlot3D for Problem 21.

In exercises 22-23, use the method of Lagrange multipliers with two constraints.

22) Optimize f(x,y,z)=yz+xy subject to the constraints: xy=1,y2+z2=1.

Answer
maximum: 32, minimum: 12

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

Answer
minimum: f(2,3,4)=29

Use the method of Lagrange multipliers to solve the following applied problems.

24) A large container in the shape of a rectangular solid must have a volume of 480 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.

25) A rectangular box without a top (a topless box) is to be made from 12 ft2 of cardboard. Find the maximum volume of such a box.

Answer
The maximum volume is 4 ft3. The dimensions are 1×2×2 ft.

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

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

Answer
(25,195)

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

Answer
1.0 unit

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

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

Answer
3 units

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

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

Answer
(1,12,3)

34) A pentagon is formed by placing an isosceles triangle on a rectangle, as shown in the diagram. If the perimeter of the pentagon is 10 in., find the lengths of the sides of the pentagon that will maximize the area of the pentagon.

CNX_Calc_Figure_14_08_201.jpg

35) [T] By investing x units of labor and y units of capital, a watch manufacturer can produce P(x,y)=50x0.4y0.6 watches. Find the maximum number of watches that can be produced on a budget of $20,000 if labor costs $100/unit and capital costs $200/unit. Use a grapher like CalcPlot3D to sketch a contour plot of the function.

Answer

Roughly 3365 watches at the critical point (80,60).

A series of curves in the first quadrant, with the first starting near (2, 120), decreasing sharply to near (20, 20), and then decreasing slowly to (120, 5). The next curve starts near (10, 120), decreases sharply to near (40, 40), and then decreases slowly to (120, 20). The next curve starts near (20, 120), decreases sharply to near (60, 60), and then decreases slowly to (120, 40). The next curve starts near (40, 120), decreases to near (80, 80), and then decreases a little slowly to (120, 60). The last curve starts near (60, 120) and decreases rather evenly through (100, 100) to (120, 90).

36) A rectangular solid is contained within a tetrahedron with vertices at (1,0,0),(0,1,0),(0,0,1), and the origin. The base of the box has dimensions x and y, and the height of the box is z. If the sum of x, y, and z is 1, find the dimensions that maximizes the volume of the rectangular solid.

37) Find the maximum value of f(x,y)=sinxsiny, where x and y denote the acute angles of a right triangle. Draw the surface plot and contour plot of the function using a CAS.

Answer

Subject to this constraint, f has a relative maximum of 12 when x=π4 and y=π4.
Surface plot and contour plot forf:

An alternating series of hills and valleys of amplitude 1 across xyz space.Contour plot for f(x, y) = sin x sin y

38) Show that, of all the triangles inscribed in a circle of radius R (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.

 

Contributors

  • Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

  • Paul Seeburger (Monroe Community College) reordered these problems, adding problems 3 and 10 and answers for problems 15 and 17. He also added a full worked-out solution for problem 9 and a link to CalcPlot3D in problems 15, 21 and 35. He also created new images for Problem 37 and expanded the answers for many problems.

 


Exercises for Lagrange Multipliers is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.

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