Ask Your Question

Revision history [back]

click to hide/show revision 1
initial version

answered 2020-09-08 10:19:52 +0200

Emmanuel Charpentier gravatar image

You have two types of conditions:

  • $a>0$, $b>0$ : those are assumptions on the parameters, not part of a condition on candidate solutions

  • $x>0$, $y>0$, $y<=x$ : those are restrictions imposed to candidate solutions.

Therefore :

# Problem-specific setup
var("x,y,a,b", domain="real")
f(x,y)=a*(x*y-y^2)*e^(-b*x^2)*e^(b*x*y)
assume(a>0,b>0)
# Extra conditions
Conds=[x>0, y>0, y<=x]
# Derivatives : must be all null to have an extremum/saddle point
Ders=[f(*f.arguments()).diff(u) for u in f.arguments()]
# Second derivatives : must be all negative to have a maximum
SDers=[f(*f.arguments()).diff(u,2) for u in f.arguments()]
# Extrema/saddle points
Exts=solve(Ders,[x,y], to_poly_solve="force", solution_dict=True)
# Candidate solutions = maxima among extrema
CSols=[u for u in Exts if all([bool(s.subs(u)<=0) for s in SDers])]
# Solutions : maxima fulfilling extra conditions
Sols=[u for u in CSols if all([v.subs(u) for v in Conds])]

sage: Sols
[]

Note that

sage: CSols
[{x: 0, y: 0}]

The (only) maximum $f$ over $\mathbb{R}^2$ is excluded by your restrictions.

HTH,

You have two types of conditions:

  • $a>0$, $b>0$ : those are assumptions on the parameters, not part of a condition on candidate solutions

  • $x>0$, $y>0$, $y<=x$ : those are restrictions imposed to candidate solutions.

Therefore :

# Problem-specific setup
var("x,y,a,b", domain="real")
f(x,y)=a*(x*y-y^2)*e^(-b*x^2)*e^(b*x*y)
assume(a>0,b>0)
# Extra conditions
Conds=[x>0, y>0, y<=x]
# Derivatives : must be all null to have an extremum/saddle point
Ders=[f(*f.arguments()).diff(u) for u in f.arguments()]
# Second derivatives : must be all negative to have a maximum
SDers=[f(*f.arguments()).diff(u,2) for u in f.arguments()]
# Extrema/saddle points
Exts=solve(Ders,[x,y], to_poly_solve="force", solution_dict=True)
# Candidate solutions = maxima among extrema
CSols=[u for u in Exts if all([bool(s.subs(u)<=0) for s in SDers])]
# Solutions : maxima fulfilling extra conditions
Sols=[u for u in CSols if all([v.subs(u) for v in Conds])]

sage: Sols
[]

Note that

sage: CSols
[{x: 0, y: 0}]

The (only) maximum $f$ over $\mathbb{R}^2$ is excluded by your restrictions.

HTH,

EDIT : I've found an easier way yo show this. Consider your problem in polar coordinates, or, more lazily, by posing y=k*x (i. e. k=arctan(theta)).

sage: var("a, b, x, y, k", domain="real")
(a, b, x, y, k)
sage: f(x, y) = a * (x * y - y^2) * exp( -b * x^2) * exp(b * x * y )

We can get rid of the special cases of the axes by noting :

sage: f(x,0)
0
sage: f(0,y)
-a*y^2

The former is constant, the latter is maximal (and zero) in y=0.

Now, what is the maximum of f along the line of equation y=k*x ?

sage: Sx=f(x,k*x).diff(x).solve(x);Sx
[x == -sqrt(-1/(b*k - b)), x == sqrt(-1/(b*k - b)), x == 0]

$$[x == -sqrt(-1/(bk - b)), x == sqrt(-1/(bk - b)), x == 0]$$

(The third case has already been disposed of...). Now, what are the values of k maximizing f . Well, the derivatives of these maxima with respect to k :

sage: Dk=[f(x,k*x).subs(u).diff(k).simplify_full() for u in Sx[:2]];Dk
[a*e^(-k/(k - 1) + 1/(k - 1))/b, a*e^(-k/(k - 1) + 1/(k - 1))/b]

$$ \left[frac{a e^{\left(-frac{k}{k - 1} + frac{1}{k - 1}\right)}}{b}, frac{a e^{\left(-frac{k}{k - 1} + frac{1}{k - 1}\right)}}{b}\right] $$

are constant :

sage: [u.log().expand_log().factor().exp().simplify_log() for u in Dk]
[a*e^(-1)/b, a*e^(-1)/b]

$$ \left[frac{a e^{\left(-1\right)}}{b}, frac{a e^{\left(-1\right)}}{b}\right] $$

Therefore no bloody maximum, notwithstanding the numerical phantasms of Sage (or Mathematica, for that matter...).

HTH,

You have two types of conditions:

  • $a>0$, $b>0$ : those are assumptions on the parameters, not part of a condition on candidate solutions

  • $x>0$, $y>0$, $y<=x$ : those are restrictions imposed to candidate solutions.

Therefore :

# Problem-specific setup
var("x,y,a,b", domain="real")
f(x,y)=a*(x*y-y^2)*e^(-b*x^2)*e^(b*x*y)
assume(a>0,b>0)
# Extra conditions
Conds=[x>0, y>0, y<=x]
# Derivatives : must be all null to have an extremum/saddle point
Ders=[f(*f.arguments()).diff(u) for u in f.arguments()]
# Second derivatives : must be all negative to have a maximum
SDers=[f(*f.arguments()).diff(u,2) for u in f.arguments()]
# Extrema/saddle points
Exts=solve(Ders,[x,y], to_poly_solve="force", solution_dict=True)
# Candidate solutions = maxima among extrema
CSols=[u for u in Exts if all([bool(s.subs(u)<=0) for s in SDers])]
# Solutions : maxima fulfilling extra conditions
Sols=[u for u in CSols if all([v.subs(u) for v in Conds])]

sage: Sols
[]

Note that

sage: CSols
[{x: 0, y: 0}]

The (only) maximum $f$ over $\mathbb{R}^2$ is excluded by your restrictions.

HTH,

EDIT : I've found an easier way yo show this. Consider your problem in polar coordinates, or, more lazily, by posing y=k*x (i. e. k=arctan(theta)).

sage: var("a, b, x, y, k", domain="real")
(a, b, x, y, k)
sage: f(x, y) = a * (x * y - y^2) * exp( -b * x^2) * exp(b * x * y )

We can get rid of the special cases of the axes by noting :

sage: f(x,0)
0
sage: f(0,y)
-a*y^2

The former is constant, the latter is maximal (and zero) in y=0.

Now, what is the maximum of f along the line of equation y=k*x ?

sage: Sx=f(x,k*x).diff(x).solve(x);Sx
[x == -sqrt(-1/(b*k - b)), x == sqrt(-1/(b*k - b)), x == 0]

$$[x == -sqrt(-1/(bk - b)), x == sqrt(-1/(bk - b)), x == 0]$$

(The third case has already been disposed of...). Now, what are the values of k maximizing f . Well, the derivatives of these maxima with respect to k :

sage: Dk=[f(x,k*x).subs(u).diff(k).simplify_full() for u in Sx[:2]];Dk
[a*e^(-k/(k - 1) + 1/(k - 1))/b, a*e^(-k/(k - 1) + 1/(k - 1))/b]

$$ \left[frac{a e^{\left(-frac{k}{k - 1} + frac{1}{k - 1}\right)}}{b}, frac{a e^{\left(-frac{k}{k - 1} + frac{1}{k - 1}\right)}}{b}\right] $$

are constant :

sage: [u.log().expand_log().factor().exp().simplify_log() for u in Dk]
[a*e^(-1)/b, a*e^(-1)/b]

$$ \left[frac{a \left[ frac{a e^{\left(-1\right)}}{b}, frac{a e^{\left(-1\right)}}{b}\right] $$

Therefore no bloody maximum, notwithstanding the numerical phantasms of Sage (or Mathematica, for that matter...).

HTH,

You have two types of conditions:

  • $a>0$, $b>0$ : those are assumptions on the parameters, not part of a condition on candidate solutions

  • $x>0$, $y>0$, $y<=x$ : those are restrictions imposed to candidate solutions.

Therefore :

# Problem-specific setup
var("x,y,a,b", domain="real")
f(x,y)=a*(x*y-y^2)*e^(-b*x^2)*e^(b*x*y)
assume(a>0,b>0)
# Extra conditions
Conds=[x>0, y>0, y<=x]
# Derivatives : must be all null to have an extremum/saddle point
Ders=[f(*f.arguments()).diff(u) for u in f.arguments()]
# Second derivatives : must be all negative to have a maximum
SDers=[f(*f.arguments()).diff(u,2) for u in f.arguments()]
# Extrema/saddle points
Exts=solve(Ders,[x,y], to_poly_solve="force", solution_dict=True)
# Candidate solutions = maxima among extrema
CSols=[u for u in Exts if all([bool(s.subs(u)<=0) for s in SDers])]
# Solutions : maxima fulfilling extra conditions
Sols=[u for u in CSols if all([v.subs(u) for v in Conds])]

sage: Sols
[]

Note that

sage: CSols
[{x: 0, y: 0}]

The (only) maximum $f$ over $\mathbb{R}^2$ is excluded by your restrictions.

HTH,

EDIT : I've found an easier way yo show this. Consider your problem in polar coordinates, or, more lazily, by posing y=k*x (i. e. k=arctan(theta)).

sage: var("a, b, x, y, k", domain="real")
(a, b, x, y, k)
sage: f(x, y) = a * (x * y - y^2) * exp( -b * x^2) * exp(b * x * y )

We can get rid of the special cases of the axes by noting :

sage: f(x,0)
0
sage: f(0,y)
-a*y^2

The former is constant, the latter is maximal (and zero) in y=0.

Now, what is the maximum of f along the line of equation y=k*x ?

sage: Sx=f(x,k*x).diff(x).solve(x);Sx
[x == -sqrt(-1/(b*k - b)), x == sqrt(-1/(b*k - b)), x == 0]

$$[x == -sqrt(-1/(bk - b)), x == sqrt(-1/(bk - b)), x == 0]$$

(The third case has already been disposed of...). Now, what are the values of k maximizing f . Well, the derivatives of these maxima with respect to k :

sage: Dk=[f(x,k*x).subs(u).diff(k).simplify_full() for u in Sx[:2]];Dk
[a*e^(-k/(k - 1) + 1/(k - 1))/b, a*e^(-k/(k - 1) + 1/(k - 1))/b]

$$ \left[frac{a e^{\left(-frac{k}{k - 1} + frac{1}{k - 1}\right)}}{b}, frac{a e^{\left(-frac{k}{k - 1} + frac{1}{k - 1}\right)}}{b}\right] $$

are constant :

sage: [u.log().expand_log().factor().exp().simplify_log() for u in Dk]
[a*e^(-1)/b, a*e^(-1)/b]

$$ \left[ frac{a e^{\left(-1\right)}}{b}, frac{a e^{\left(-1\right)}}{b}\right] $$

Therefore no bloody maximum, notwithstanding the numerical phantasms of Sage (or Mathematica, for that matter...).

HTH,

You have two types of conditions:

  • $a>0$, $b>0$ : those are assumptions on the parameters, not part of a condition on candidate solutions

  • $x>0$, $y>0$, $y<=x$ : those are restrictions imposed to candidate solutions.

Therefore :

# Problem-specific setup
var("x,y,a,b", domain="real")
f(x,y)=a*(x*y-y^2)*e^(-b*x^2)*e^(b*x*y)
assume(a>0,b>0)
# Extra conditions
Conds=[x>0, y>0, y<=x]
# Derivatives : must be all null to have an extremum/saddle point
Ders=[f(*f.arguments()).diff(u) for u in f.arguments()]
# Second derivatives : must be all negative to have a maximum
SDers=[f(*f.arguments()).diff(u,2) for u in f.arguments()]
# Extrema/saddle points
Exts=solve(Ders,[x,y], to_poly_solve="force", solution_dict=True)
# Candidate solutions = maxima among extrema
CSols=[u for u in Exts if all([bool(s.subs(u)<=0) for s in SDers])]
# Solutions : maxima fulfilling extra conditions
Sols=[u for u in CSols if all([v.subs(u) for v in Conds])]

sage: Sols
[]

Note that

sage: CSols
[{x: 0, y: 0}]

The (only) maximum $f$ over $\mathbb{R}^2$ is excluded by your restrictions.

HTH,

EDIT : I've found an easier way yo show this. Consider your problem in polar coordinates, or, more lazily, by posing y=k*x (i. e. k=arctan(theta)).

sage: var("a, b, x, y, k", domain="real")
(a, b, x, y, k)
sage: f(x, y) = a * (x * y - y^2) * exp( -b * x^2) * exp(b * x * y )

We can get rid of the special cases of the axes by noting :

sage: f(x,0)
0
sage: f(0,y)
-a*y^2

The former is constant, the latter is maximal (and zero) in y=0.

Now, what is the maximum of f along the line of equation y=k*x ?

sage: Sx=f(x,k*x).diff(x).solve(x);Sx
[x == -sqrt(-1/(b*k - b)), x == sqrt(-1/(b*k - b)), x == 0]

$$[x == -sqrt(-1/(bk - b)), x == sqrt(-1/(bk - b)), x == 0]$$

(The third case has already been disposed of...). Now, what are the values of k maximizing f . Well, the derivatives of these maxima with respect to k :

sage: Dk=[f(x,k*x).subs(u).diff(k).simplify_full() for u in Sx[:2]];Dk
[a*e^(-k/(k - 1) + 1/(k - 1))/b, a*e^(-k/(k - 1) + 1/(k - 1))/b]

are constant :

sage: [u.log().expand_log().factor().exp().simplify_log() for u in Dk]
[a*e^(-1)/b, a*e^(-1)/b]

Therefore no bloody maximum, notwithstanding the numerical phantasms of Sage (or Mathematica, for that matter...).

HTH,

Copyright Sage, 2010. Some rights reserved under creative commons license. Content on this site is licensed under a Creative Commons Attribution Share Alike 3.0 license.
about | faq | help | privacy policy | terms of service
Powered by Askbot version 0.7.59