# Revision history [back]

One week after, we stille don't know if this is homework or not. Well... one week after, it might be stale. So here are a couple of answers to a question interestng beyond the OP potential homework.

The problem data can be set up in sage as follows :

# Problem data
# Arguments of the function to optimize and nullity constraints to satisfy.
Args = var("w, x, y, z")
# Function tooptimize (here a Symbolic Expression)
F = (4*x + 4*y + 9*z - 2*w)
# Constraints (here, symbolic expressions to nullify)
G = (2*x + 2*y + z + w)
H = (x^2 + y^2 + z - 4)
# Problem setup
# Tuple of constraints :
# we no longer depend of the particular shape of the problem
C = (G, H)


### General solution

The general solution states that if (the$n$-dimensional vector) $\mathbf{x^}$ maximizes (the scalar) $\phi(\mathbf{x})$ under the ($m$-vector) *equality constraint(s) $\mathbf{\psi(x)}=\mathbf{0}$, then there exist an ($m$-dimensional) vector $\mathbf{\lambda}$ (the Lagrange multipliers) such as all the components of the differential of the (scalar) function (callad Lagrangian) $\phi(\mathbf{x})+ \mathbf{\lambda}\cdot\mathbf{\psi(c)}$ are all null ; in other words, the maximization of a scalar functin of $n$ variables under $m$ constraints is replaced by the maximization of a scalar function of $n+m$ variables.

This translates in Sage as :

# Solution
# (Vector) function of constraints
psi = vector(C).function(*Args)
# Vector of Lagrange multipliers
Lambda = vector([var("lambda_{}".format(u)) for u in range(len(C))])
# Function to optimize
phi = F.function(*Args)
# Lagrangian arguments
ArgsL = Args + tuple(Lambda)
# Lagrangian function
L = (phi + Lambda.dot_product(psi)).function(*ArgsL)
# Its differential (vector)
dL = L.diff()(*ArgsL)
# Solution(s) nullifying this differential :
Sol = solve(list(dL), ArgsL, solution_dict=True)
# Extremal value
Max = F.subs(Sol)


which gives :

sage: Sol
[{w: -628/121, x: 4/11, y: 4/11, z: 452/121, lambda_0: 2, lambda_1: -11}]
sage: Max
516/11


Nota bene : the Lagrange multipliers solution(s) can be maxima or minima, but may also be saddle points ; this should be checked.

### Specialized solution

The idea is to use the $m$ constraints to express $m$ variables as functins of the other $n-m$ variables. (Here, we can express $w$ and $z$ as finctions of $x$ and $y$). The problem is then to fonfd the (unconstrained) maximum of a finction of $n-m$ variables, wich is done by solving the components of its differential for tjhose variables :

# Eliminate two variables from the two constraints. Here, we choose w and z,
# since one constraint uses x^2 and y^2, opening the door to multiple solutions.
S1=solve([G, H],[w, z], solution_dict=True)
# Substitute this in F
F1 = F.subs(S1)
# Optimize F1, i. e. find the point(s) where both derivatives are null :
Sol1 = solve([F1.diff(u) for u in (x, y)],[x, y], solution_dict=True)
# Substitute that in the [G, H] system, and solve for (numerical) w, z :
Sol2 = solve([u.subs(Sol1[0]) for u in (G, H)],[w, z], solution_dict=True)
# Cleanup : One dictionary solution :
HSSol = copy(Sol1[0])
HSSol.update(Sol2[0])
# Get the (numerical) extremum :
HSMax = F.subs(HSSol)


which gives :

sage: HSSol
{x: 4/11, y: 4/11, w: -628/121, z: 452/121}
sage: HSMax
516/11


Again, the nature (maximum, minimum or saddle-point) of the extrema thus obtained must be checked.

### Numerical solutions

One can note that Sage implements numerical methods of minimization of a real function subject to (more general inequality) constraints, accessible via the minimize_constrained function. Other methods implemented in Scipy (which is part of Sage) can also be used.

 2 No.2 Revision slelievre 15169 ●16 ●139 ●302 http://carva.org/samue...

One week after, we stille still don't know if this is homework or not. not. Well... one week after, it might be stale. So here are a couple of of answers to a question interestng interesting beyond the OP OP's potential homework.

The problem data can be set up in sage Sage as follows :follows:

# Problem data
# Arguments of the function to optimize and nullity constraints to satisfy.
satisfy
Args = var("w, x, y, z")
# Function tooptimize to optimize (here a Symbolic Expression)
symbolic expression)
F = (4*x + 4*y + 9*z - 2*w)
# Constraints (here, symbolic expressions to nullify)
G = (2*x + 2*y + z + w)
H = (x^2 + y^2 + z - 4)
# Problem setup
# Tuple setup: tuple of constraints :
# we no longer -- expressed in
# a way that does not depend of on the particular shape of the problem
C = (G, H)


The general general solution states that if (the$n$-dimensional (the $n$-dimensional vector) $\mathbf{x^}$ $\mathbf{x^{*}}$ maximizes (the scalar) $\phi(\mathbf{x})$ under the ($m$-vector) *equality constraint(s) $\mathbf{\psi(x)}=\mathbf{0}$, equality constraint $\mathbf{\psi(x)} = \mathbf{0}$, then there exist an ($m$-dimensional) vector vector $\mathbf{\lambda}$ (the Lagrange multipliers) such as all the components of the the differential of the (scalar) function (callad Lagrangian) Lagrangian) $\phi(\mathbf{x})+ \mathbf{\lambda}\cdot\mathbf{\psi(c)}$ \mathbf{\lambda} \cdot \mathbf{\psi(c)}$are all null ; zero; in other words, the maximization of a scalar functin of$n$variables variables under$m$constraints is replaced by the maximization maximization of a scalar function of$n+mn + m$variables. This translates in Sage as :as: # Solution # (Vector) function of constraints psi = vector(C).function(*Args) # Vector of Lagrange multipliers Lambda = vector([var("lambda_{}".format(u)) for u in range(len(C))]) # Function to optimize phi = F.function(*Args) # Lagrangian arguments ArgsL = Args + tuple(Lambda) # Lagrangian function L = (phi + Lambda.dot_product(psi)).function(*ArgsL) # Its differential (vector) dL = L.diff()(*ArgsL) # Solution(s) nullifying this differential : Sol = solve(list(dL), ArgsL, solution_dict=True) # Extremal value Max = F.subs(Sol)  which gives :gives: sage: Sol [{w: -628/121, x: 4/11, y: 4/11, z: 452/121, lambda_0: 2, lambda_1: -11}] sage: Max 516/11  Nota bene :bene: the Lagrange multipliers solution(s) solutions can be maxima or minima, minima, but may also be saddle points ; points; this should be checked. ### Specialized solution The idea is to use the$m$constraints to express$m$variables as functins of functions of the other$n-mn - m$variables. (Here, we can express$w$and$z$as finctions functions of$x$and$y$). The problem is then to fonfd find the (unconstrained) maximum of a finction of$n-m$maximum of a function of$n - m$variables, wich which is done by solving the components components of its differential for tjhose variables :those variables: # Eliminate two variables from the two constraints. Here, we choose w and z, # since one constraint uses x^2 and y^2, opening the door to multiple solutions. S1=solve([G, H],[w, solutions S1 = solve([G, H], [w, z], solution_dict=True) # Substitute this in F F1 = F.subs(S1) # Optimize F1, i. e. find the point(s) where both derivatives are null : zero Sol1 = solve([F1.diff(u) for u in (x, y)],[x, y)], [x, y], solution_dict=True) # Substitute that in the [G, H] system, and solve for (numerical) w, z : z Sol2 = solve([u.subs(Sol1[0]) for u in (G, H)],[w, H)], [w, z], solution_dict=True) # Cleanup : # Cleanup: One dictionary solution : solution HSSol = copy(Sol1[0]) HSSol.update(Sol2[0]) # Get the (numerical) extremum : extremum HSMax = F.subs(HSSol)  which gives :gives: sage: HSSol {x: 4/11, y: 4/11, w: -628/121, z: 452/121} sage: HSMax 516/11  Again, the nature (maximum, minimum or saddle-point) saddle-point) of the extrema thus obtained must be checked. ### Numerical solutions One can note that Sage implements numerical methods of minimization minimization of a real function subject to (more general inequality) constraints, inequality) constraints, accessible via the minimize_constrained function. Other methods methods implemented in Scipy SciPy (which is part of Sage) Sage depends on) can also be used. One week after, we still don't know if this is homework or not. Well... one week after, it might be stale. So here are a couple of answers to a question interesting beyond the OP's potential homework. The problem data can be set up in Sage as follows: # Problem data # Arguments of the function to optimize and nullity constraints to satisfy Args = var("w, x, y, z") # Function to optimize (here a symbolic expression) F = (4*x + 4*y + 9*z - 2*w) # Constraints (here, symbolic expressions to nullify) G = (2*x + 2*y + z + w) H = (x^2 + y^2 + z - 4) # Problem setup: tuple of constraints -- expressed in # a way that does not depend on the particular problem C = (G, H)  ### General solution The general solution states that if (the$n$-dimensional vector)$\mathbf{x^{*}}\mathbf{x_0}$maximizes (the scalar)$\phi(\mathbf{x})$under the ($m$-vector) equality constraint$\mathbf{\psi(x)} $\boldsymbol{\psi}\mathbf{(x}) = \mathbf{0}$, then there exist an ($m$-dimensional) vector $\mathbf{\lambda}$ (the Lagrange multipliers) such as all the components of the differential of the (scalar) function (callad Lagrangian) $\phi(\mathbf{x})+$\phi(\mathbf{x_0})+ \mathbf{\lambda} \cdot \mathbf{\psi(c)}$\mathbf{\boldsymbol{\psi}(x_0)}$ are all zero; in other words, the maximization of a scalar functin function of $n$ variables under $m$ constraints is replaced by the maximization of a scalar function of $n + m$ variables.

This translates in Sage as:

# Solution
# (Vector) function of constraints
psi = vector(C).function(*Args)

# Vector of Lagrange multipliers
Lambda = vector([var("lambda_{}".format(u)) for u in range(len(C))])

# Function to optimize
phi = F.function(*Args)

# Lagrangian arguments
ArgsL = Args + tuple(Lambda)

# Lagrangian function
L = (phi + Lambda.dot_product(psi)).function(*ArgsL)

# Its differential (vector)
dL = L.diff()(*ArgsL)

# Solution(s) nullifying this differential :
Sol = solve(list(dL), ArgsL, solution_dict=True)

# Extremal value
Max = F.subs(Sol)


which gives:

sage: Sol
[{w: -628/121, x: 4/11, y: 4/11, z: 452/121, lambda_0: 2, lambda_1: -11}]
sage: Max
516/11


Nota bene: the Lagrange multipliers solutions can be maxima or minima, but may also be saddle points; this should be checked.

### Specialized solution

The idea is to use the $m$ constraints to express $m$ variables as functions of the other $n - m$ variables. (Here, we can express $w$ and $z$ as functions of $x$ and $y$). The problem is then to find the (unconstrained) maximum of a function of $n - m$ variables, which is done by solving the components of its differential for those variables:variables.

This is called parameterization.

# Eliminate two variables from the two constraints. Here, we choose w and z,
# since one constraint uses x^2 and y^2, opening the door to multiple solutions
S1 = solve([G, H], [w, z], solution_dict=True)

# Substitute this in F
F1 = F.subs(S1)

# Optimize F1, i. e. find the point(s) where both derivatives are zero
Sol1 = solve([F1.diff(u) for u in (x, y)], [x, y], solution_dict=True)

# Substitute in the [G, H] system, and solve for (numerical) w, z
Sol2 = solve([u.subs(Sol1[0]) for u in (G, H)], [w, z], solution_dict=True)

# Cleanup: One dictionary solution
HSSol = copy(Sol1[0])
HSSol.update(Sol2[0])

# Get the (numerical) extremum
HSMax = F.subs(HSSol)


which gives:

sage: HSSol
{x: 4/11, y: 4/11, w: -628/121, z: 452/121}
sage: HSMax
516/11


Again, the nature (maximum, minimum or saddle-point) of the extrema thus obtained must be checked.

### Numerical solutions

One can note that Sage implements numerical methods of minimization of a real function subject to (more general inequality) constraints, accessible via the minimize_constrained function. Other methods implemented in SciPy (which Sage depends on) can also be used.