Information for Maple assignment 5

Each student will get individual data for the assignment.

Lagrange multipliers using Maple

Background

Here is the problem:

Theoretical results imply that x+3yz has a maximum and a minimum on the sphere x2+y2+z2=1. Use Lagrange multipliers to find these maximum and minimum values.

The constraint in this problem is x2+y2+z2=1, a sphere which easy to understand and to visualize. The objective function is x+3yz. If k>0, then the points in R3 which satisfy x+kyz=constant form a graph which is a hyperboloid with one sheet. We will consider first what happens if k=3, the problem stated above.

A level surface of the objective function together with the constraint surface

Here is a picture of x+3yz=.5 and x2+y2+z2=1. The .5 is a random value of the constant. This hyperboloid divides the sphere into two pieces. On one piece of the sphere, the function x+3yz has values larger than .5, and on the other piece, the values of the function are less than .5.
The picture uses display3d with ϑ=80 and φ=−75 to show two graphs combined. The graphs are the results of the following implicit3d commands for the sphere and hyperboloid respectively. The numpoints=1800 option increases the number of points sampled from the default of 1000. With only 1000 points, the graphs drawn for the surfaces look very polyhedral. That is, they resemble more closely linear approximations to the true smooth graphs.

implicitplot3d(x^2+y^2+z^2=1,x=-2..2,y=-2..2,z=-2..2,axes=normal,numpoints=1800,color=blue);
implicitplot3d(x+3*y*z=.5,x=-2..2,y=-2..2,z=-2..2,axes=normal,numpoints=1800,color=green);

Lagrange multipliers and extreme values

Let's consider the surface f(x,y,z)=x+3yz=constant and its intersection with the constraint surface, the sphere g(x,y,z)=x2+y2+z2=1. If the surfaces are not tangent at an intersection point (such surfaces are called transverse), then there will be values of f greater and less than constant on the constraint surface. This will not happen if the surfaces are tangent. The Lagrange multiplier equations follow from this observation. We therefore look for extreme values among the solutions x, y, and z of the equations
   ∇f(x,y,z)=λ∇g(x,y,z)
Here λ is a real number, and the equality indicates that the tangent planes are parallel because the normal vectors are pointing in the same direction. This one vector equation is three scalar equations.
   g(x,y,z)=1
This guarantees that the solutions we seek will be on the constraint surface.

In this specific case, we want to solve these equations
   1= λ(2x)
   3z=λ(2y)
   3y=λ(2z)
   1=x2+y2+z2

Finding solutions using Maple

Here is one way to get solutions. 
> f:=x+3*y*z;

                                f := x + 3 y z

> g:=x^2+y^2+z^2;

                              2    2    2
                        g := x  + y  + z

> eqx:=diff(f,x)=L*diff(g,x);

                      eqx := 1 = 2 L x

> eqy:=diff(f,y)=L*diff(g,y);

                      eqy := 3 z = 2 L y

> eqz:=diff(f,z)=L*diff(g,z);

                      eqz := 3 y = 2 L z

> sols:=solve({eqx,eqy,eqz,g=1});

sols := {y = 0, x = 1, L = 1/2, z = 0}, {y = 0, x = -1, L = -1/2, z = 0},

            {L = 3/2, x = 1/3, y = 2/3, z = 2/3},

           {L = 3/2, x = 1/3, y = -2/3, z = -2/3},

           {L = -3/2, x = -1/3, y = 2/3, z = -2/3},

           {L = -3/2, x = -1/3, y = -2/3, z = 2/3}

> values:=seq(subs(sols[j],x+3*y*z),j=1..nops({sols}));

                        values := 1, -1, 5/3, 5/3, -5/3, -5/3

Comments on the Maple commands and a picture

The command eqx:=diff(f,x)=L*diff(g,x) labels the equation ∂f/∂x=L(∂g/∂x) as eqx so that it can be referred to later. The = is part of an equation, and the symbols := are used by Maple to assign a value. In this case, the "label" eqx is assigned as its value the equation diff(f,x)=L*diff(g,x).

L is used instead of λ. The solve step finds all solutions of the system of equations. Finally, the command values:=... substitutes the values found into the constraint function. Inside, the nops({sols}) instructs Maple to count the number of elements in the set of solutions -- that is, the number of solutions to the Lagrange multiplier equations. Then Maple will substitute each of these solutions, in order, into the constraint function.

Therefore we have learned that the maximum value of the objective function is 5/3. This value is attained at exactly two points: (1/3,2/3,2/3) and(1/3,–2/3,–2/3). The minimum value of the objective function is –5/3, and this is attained at the two points (–1/3,2/3,–2/3) and (–1/3,–2/3,2/3).

A picture of the constraint surface with the level surface corresponding to constant=5/3 is shown to the right. The two points of tangency shown are the points found by the Lagrange multiplier method.

A common error ...

Students working by hand frequently discard two of the solutions found by Maple. These are the solutions with y=0 and z=0 -- perhaps these are thought to be "trivial" solutions which can't give useful answers. Such solutions should not be thrown away. For the objective function x+3xy these solutions are not extreme values but are more complicated.

A graph of the constraint and the level surface x+3xy=1 near the point (1,0,0) is shown to the right. The window was chosen with some care, since the features here are quite delicate: x is between .975 and 1.025 and y and z are both in the interval [−.15,.15]. The two surfaces are tangent, but the constraint surface is locally both inside and outside of the sphere -- that is, the function x+3yz has both larger and smaller values than constant=1 on x2+y2+z2=1 near (1,0,0). The objective function when restricted to the constraint surface near (1,0,0) therefore has saddle behavior.

     Change the objective function slightly:
don't throw things away, but be careful!

Suppose we keep the unit sphere x2+y2+z2=1 as constraint, and consider the objective function f(x,y,z)=x+.5yz (the 1 in front of yz has been changed to .5). What follows is true for x+kyz if 0≤k≤1, but keeping track of the results for one value of k will be complicated enough!

The maximum value of this f on the unit sphere is 1, and this value occurs only at the point (1,0,0). The minimum value is −1, which occurs only at the point (−1,0,0). A picture of the constraint surface and the level surface of the objective function which corresponds to the maximum value, the graph of x+.5yz=1, is shown to the right.

Algebraically, this is a solution to the Lagrange multiplier equations which has both y=0 and z=0. This sort of solution is exactly one which may be thrown away when people compute "by hand".

Here is a sequence of Maple commands similar to what was just done. We only display the results of the last command. This should give the values of the objective function which result. The extreme values should be among these "candidates". 

> f:=x+.5*y*z:

> g:=x^2+y^2+z^2:

> eqx:=diff(f,x)=L*diff(g,x):

> eqy:=diff(f,y)=L*diff(g,y):

> eqz:=diff(f,z)=L*diff(g,z):

> sols:=solve({eqx,eqy,eqz,g=1}):

> values:=seq(subs(sols[j],x+.5*y*z),j=1..nops({sols}));

      values := 1., -1., 1.250000000, 1.250000000, -1.250000000, -1.250000000}
The values which Maple returns as candidates for extreme values are ±1 and ±1.25.

To the right is a graph of the unit sphere and the surface x+.5yz=1.25, at an angle similar to what was just shown before. The surface for the candidate "extreme value" does not seem to touch the unit sphere! What happened?

Maple is a computer program. It will try to do exactly the computations requested by the user.

    Be careful!

Let's show a bit more information about by looking in detail at the result of one of the intermediate steps, the solve command. 

> sols:=solve({eqx,eqy,eqz,g=1});

        sols := {z = 0., x = 1., y = 0., L = 0.5000000000},
 
           {z = 0., y = 0., x = -1., L = -0.5000000000},

    {x = 2., L = 0.2500000000, z = 1.224744871 I, y = 1.224744871 I},

   {x = 2., L = 0.2500000000, z = -1.224744871 I, y = -1.224744871 I},

   {x = -2., z = 1.224744871 I, y = -1.224744871 I, L = -0.2500000000},

   {x = -2., y = 1.224744871 I, z = -1.224744871 I, L = -0.2500000000}
The third element of sols is
     {x = 2., L = 0.2500000000, z = 1.224744871 I, y = 1.224744871 I}
where I is Maple's notation for i, or the square root of −1. If we compute 2.+.5* 1.224744871 I* 1.224744871 I (that's x+.5yz at this "point") then the result, to 10 digit accuracy, will be 1.250000000.

Here is a quote from a Maple help page. It is an important part of the design of Maple which should be known to every user:

      By default, Maple performs computations under the assumption that the underlying number system is the complex field.

So, for example, we have: 

> solve(x^2+1);

                         I, -I
We can change this default setting. In the case of computations with Lagrange multipliers such a change is useful. Consider the following commands and responses. 
> with(RealDomain):

> solve(x^2+1);

> sols:=solve({eqx,eqy,eqz,g=1});

         sols := {z = 0., x = 1., y = 0., L = 0.5000000000},

           {z = 0., y = 0., x = -1., L = -0.5000000000}

> values:=seq(subs(sols[j],x+.5*y*z),j=1..nops({sols}));

                               values := 1., -1.
When the RealDomain package is loaded, Maple declares that x2+1=0 has no roots. It also discards any elements of sols which have non-zero imaginary parts, and so it declares as a result of the substitution step, only those values which correspond to real solutions of the Lagrange multiplier equations. We don't get the outputs ±1.25 because the intermediate steps involved imaginary numbers which RealDomain discards. Those solutions won't be geometrically valid.

Maintained by greenfie@math.rutgers.edu and last modified 3/3/2010.