Lecture Notebooks for Vector Calculus​
Lagrange multipliers
​By Passant Abbassi​
Wolfram Summer School 2022
​passant.abbassy@guc.edu.eg​
Project Mentors: Paul Abbott and Daniel Robinson, Wolfram Research

Motivation

Many problems in Engineering and science require us to find the maximum or minimum of a differentiable function f. The scope of the project is to create a lecture notebook on Lagrange multipliers. The main aim is to make good use of lecture notebooks in the classroom by including live interactive content that can be altered—and even extended, while presenting. This lets classes be truly dynamic and provides an unparalleled opportunity to involve students in the content creation.

Method of Lagrange Multipliers

Optimization with a constraint for functions of two or three variables

How do you implement Lagrange multipliers?
1
.
Compute gradient:
∇
{x,y}
f(x,y),
∇
{x,y}
g(x,y)
f
x
,
f
y
,
g
x
,
g
y
2
.
Apply lagrange multipliers:
∇
{x,y}
f(x,y)=λ
∇
{x,y}
g(x,y)⟹
f
x
=λ
g
x
,
f
y
=λ
g
y
3
.
Solve the 3 equations
f
x
=λ
g
x
,
f
y
=λ
g
y
,g(x,y)=k
and 3 unknowns
(x,y,λ)
to get the critical points
4
.
Plug the critical points in the function
f(x,y)
to identify max and min values: The point that gives largest value of function
f(x,y)
is Absolute maximum while the point that gives smallest value of function
f(x,y)
is Absolute minimum.
As an example, let’s find the extreme values of
f(x,y)
subject to the constraint
g(x,y)=k
.

Example 1: Find extreme values of
f(x,y)=
2
x
+2
2
y
subject to the constraint
g(x,y)=
2
x
+
2
y

In this example,
f(x,y)=
2
x
+2
2
y
subject to the constraint
g(x,y)=
2
x
+
2
y
.
In[]:=
f[x_,y_]:=
2
x
+2
2
y
In[]:=
g[x_,y_]:=
2
x
+
2
y
To find the maximum and minimum values of
f(x,y)
subject to the constraint
g(x,y)=k
, apply Lagrange multipliers.
Find all values of
x,y
and
λ
such that:
∇
{x,y}
f(x,y)=λ
∇
{x,y}
g(x,y)⟹
f
x
=λ
g
x
,
f
y
=λ
g
y
1
.
Compute gradient:
In[]:=
gradf=
∇
{x,y}
f[x,y]
Out[]=
{2x,4y}
In[]:=
gradg=
∇
{x,y}
g[x,y]
Out[]=
{2x,2y}
2
.
Apply Lagrange multipliers:
Convert the input to TraditionalForm:
In[]:=
∇
{x,y}
f(x,y)λ
∇
{x,y}
g(x,y)
Out[]=
{2x,4y}{2xλ,2yλ}
In[]:=
lagrange=Thread[Equal[{2x,4y},{2xλ,2yλ}]]//Column
Out[]=
2x2xλ
4y2yλ
In[]:=
(system=Flatten[{lagrange,g[x,y]1}])//Column
Out[]=
16x8xλ
4z2yλ
-16+4y8zλ
2
x
+
2
y
1
3
.
Solve the 3 equations and 3 unknowns to get the critical points:
In[]:=
Solve[system,{x,y,λ}]
Out[]=
{{x0,y-1,λ-1},{x0,y1,λ-1}}
Equation (1) implies that
x=0
or
λ=1
.
◼
  • If
    x=0
    , then equation (3) implies that
    y=±1
    . This yields 2 points
    (0,±1)
    • ◼
  • If
    λ=1
    , then equation (2) implies that
    y=0
    and by equation (3),
    x=±1
    . This yields 2 points
    (±1,0)
    • 4
    .
    Plug the critical points in the function
    f(x,y)
    to identify max and min values:
    Now evaluating
    f(x,y)
    at the 4 points
    (±1,0)
    and
    (0,±1)
    ◼
  • f(±1,0)=1
    and
    f(0,±1)=2
    • 5
    .
    Conclusion:
    ◼
  • f(x,y)
    attains its largest value of 2 (Absolute max) at either point
    (0,±1)
    • ◼
  • f(x,y)
    attains its smallest of 1 (Absolute min) at either point
    (±1,0)

    • Visualize the functions and their contour plots

    Plot of function
    f(x,y)
    :
    In[]:=
    plot3D=Plot3D[f[x,y],{x,-1.5,1.5},{y,-1.5,1.5},PlotLabelStyle["f(x,y)=
    2
    x
    +2
    2
    y
    ",18,Bold,Red]]
    Out[]=
    Contour Plot of function
    f(x,y)
    :
    In[]:=
    fplot=ContourPlot[f[x,y],{x,-1.5,1.5},{y,-1.5,1.5},PlotLabelStyle["Contour Plot of function f(x,y)",18,Bold,Red]]
    Out[]=

    Example 2: Find the minimum distance from the surface
    2
    x
    -
    2
    y
    -
    2
    z
    =1
    to the origin

    We wish to minimize the value of function
    d(x,y)=
    2
    (x-0)
    +
    2
    (y-0)
    +
    2
    (z-0)
    =
    2
    x
    +
    2
    y
    +
    2
    z
    for points
    (x,y,z)
    on the surface
    2
    x
    -
    2
    y
    -
    2
    z
    =1
    . Equivalently, the problem may be formulated as finding the minimum of squared distance from the origin
    d(x,y)=
    2
    x
    +
    2
    y
    +
    2
    z
    subject to the constraint
    g(x,y,z)=
    2
    x
    -
    2
    y
    -
    2
    z
    =1
    .
    In[]:=
    d[x_,y_,z_]:=
    2
    x
    +
    2
    y
    +
    2
    z
    In[]:=
    g[x_,y_,z_]:=
    2
    x
    -
    2
    y
    -
    2
    z
    To find the minimum values of
    d(x,y,z)
    subject to the constraint
    g(x,y,z)=k
    , apply the Lagrange multipliers.
    In other words, find all values of
    x,y,z
    and
    λ
    such that:
    ∇
    {x,y}
    f(x,y,z)=λ
    ∇
    {x,y}
    g(x,y,z)⟹
    f
    x
    =λ
    g
    x
    ,
    f
    y
    =λ
    g
    y
    ,
    f
    z
    =λ
    g
    z
    1
    .
    Compute gradient:
    In[]:=
    gradd=Grad[d[x,y,z],{x,y,z}]
    Out[]=
    {2x,2y,2z}
    In[]:=
    gradd=Grad[g[x,y,z],{x,y,z}]
    Out[]=
    {2x,-2y,-2z}
    2
    .
    Apply Lagrange multipliers:
    In[]:=
    ∇
    {x,y,z}
    d[x,y,z]λ
    ∇
    {x,y,z}
    g[x,y,z]
    Out[]=
    {2x,2y,2z}{2xλ,-2yλ,-2zλ}
    In[]:=
    lagrange=Thread[Equal[{2x,2y,2z},{2xλ,-2yλ,-2zλ}]];​​lagrange//Column
    Out[]=
    2x2xλ
    2y-2yλ
    2z-2zλ
    3
    .
    Solve the 4 equations and 4 unknowns to get the critical points:
    In[]:=
    system=Join[lagrange,{g[x,y,z]1}];​​system//Column
    Out[]=
    2x2xλ
    2y-2yλ
    2z-2zλ
    2
    x
    -
    2
    y
    -
    2
    z
    1
    
    Reduce will return a set of equations equivalent to the input. Solve will try to find a solution (it makes more assumptions).

    Example 3: Hottest point on space probe’s surface

    Find the hottest point on the probe's surface.
    1
    .
    Compute gradient:
    2
    .
    Apply Lagrange multipliers:
    3
    .
    Solve the 4 equations and 4 unknowns to get the critical points: