Finding a constraint on one variable of a multivariable function to constrain the entire function ...
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Finding a constraint on one variable of a multivariable function to constrain the entire function
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$begingroup$
I have a function.
Now I want to let my variables only take values between 0, and 1.
The problem is as follows. For what values of Y, is L(x,y) < 0.
That is, without putting a further constraint on x(other than its between 0, 1), how can you bind Y, such that you bind L(x,y) below 0.
I graphed it out, and there is visually a range of y such that this is true.
I was unable to proceed. I thought of trying the mins and max for x, but then I noticed that really doesn't give me any useful information. I understand that the last inequality must be true for whenever L<0, including the range im looking for, but there must be some other point I am not taking into account that will give me a range independent of x.
multivariable-calculus optimization lagrange-multiplier maxima-minima constraints
asked Mar 21 at 3:54
physicsPphysicsP
386
$endgroup$
add a comment |
$begingroup$
I have a function.
Now I want to let my variables only take values between 0, and 1.
The problem is as follows. For what values of Y, is L(x,y) < 0.
That is, without putting a further constraint on x(other than its between 0, 1), how can you bind Y, such that you bind L(x,y) below 0.
I graphed it out, and there is visually a range of y such that this is true.
I was unable to proceed. I thought of trying the mins and max for x, but then I noticed that really doesn't give me any useful information. I understand that the last inequality must be true for whenever L<0, including the range im looking for, but there must be some other point I am not taking into account that will give me a range independent of x.
multivariable-calculus optimization lagrange-multiplier maxima-minima constraints
asked Mar 21 at 3:54
physicsPphysicsP
386
$endgroup$
add a comment |
$begingroup$
I have a function.
Now I want to let my variables only take values between 0, and 1.
The problem is as follows. For what values of Y, is L(x,y) < 0.
That is, without putting a further constraint on x(other than its between 0, 1), how can you bind Y, such that you bind L(x,y) below 0.
I graphed it out, and there is visually a range of y such that this is true.
I was unable to proceed. I thought of trying the mins and max for x, but then I noticed that really doesn't give me any useful information. I understand that the last inequality must be true for whenever L<0, including the range im looking for, but there must be some other point I am not taking into account that will give me a range independent of x.
multivariable-calculus optimization lagrange-multiplier maxima-minima constraints
asked Mar 21 at 3:54
physicsPphysicsP
386
$endgroup$
I have a function.
Now I want to let my variables only take values between 0, and 1.
The problem is as follows. For what values of Y, is L(x,y) < 0.
That is, without putting a further constraint on x(other than its between 0, 1), how can you bind Y, such that you bind L(x,y) below 0.
I graphed it out, and there is visually a range of y such that this is true.
I was unable to proceed. I thought of trying the mins and max for x, but then I noticed that really doesn't give me any useful information. I understand that the last inequality must be true for whenever L<0, including the range im looking for, but there must be some other point I am not taking into account that will give me a range independent of x.
multivariable-calculus optimization lagrange-multiplier maxima-minima constraints
multivariable-calculus optimization lagrange-multiplier maxima-minima constraints
asked Mar 21 at 3:54
physicsPphysicsP
386
asked Mar 21 at 3:54
physicsPphysicsP
386
edited Mar 23 at 19:38
physicsP
asked Mar 21 at 3:54
physicsPphysicsP
386
asked Mar 21 at 3:54
physicsPphysicsP
386
asked Mar 21 at 3:54
physicsPphysicsP
386
386
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider the four optimization problems, which cover the combinations of minimizing or maximizing y, when x is fixed at 0 or 1.
a) min w.r.t. y subject to $0 le y le 1, L(x,y) < 0, x = 0$
b) max w.r.t. y subject to $0 le y le 1, L(x,y) < 0, x = 0$
c) min w.r.t. y subject to $0 le y le 1, L(x,y) < 0, x = 1$
d) max w.r.t. y subject to $0 le y le 1, L(x,y) < 0, x = 1$
Solving these as having non-strict inequality $L(x,y) le 0$, the following solutions are obtained:
a) x = 0, y = 1/3
b) x = 0, y = 1
c) x = 1, y = 0
d) x = 1, y = 2/5.5
Given the linearity of L(x,y) for a fixed x, we can conclude that
when x = 0, then y in the interval (1/3,1] satisfies L(x,y) < 0
when x = 1, then y in the interval [0,2.5/5) satisfies L(x,y) < 0.
Due to the bilinearity of L(x,y), x = 0 and 1 are the extreme cases for x in [0,1] with regard to minimum and maximum possible values of y satisfying L(x,y) < 0.
Noting that 2.5/5 = 0.3626... we can therefore conclude that every value of y in the interval (1/3,2.5/5), and only values of y in that interval, will satisfy L(x,y) < 0 no matter which (i.e., "worst case") value of x in [0,1] is chosen.
answered Mar 23 at 17:42
Mark L. StoneMark L. Stone
1,95058
$endgroup$
$begingroup$
What do you mean when you say linearity? "Given the linearity of L(x,y) for a fixed x" do you mean for a fixed x it is linear in x or y? What exactly is the reasoning behind using edges?
$endgroup$
– physicsP
Mar 23 at 19:28
1
$begingroup$
For a fixed x, L(x,y) is linear in y.
$endgroup$
– Mark L. Stone
Mar 23 at 19:29
$begingroup$
What do you mean by "using edges" in your comment question "What exactly is the reasoning behind using edges?"? Are you asking about my use of the term "extreme cases".? I chose x = 0 and x = 1 as extreme cases, because no intermediate value of x could lead to more extreme bounds on possible y because L(x,y) is linear in y for a fixed x.
$endgroup$
– Mark L. Stone
Mar 23 at 19:44
$begingroup$
You have now changed the question so as to eliminate the specification of particular function L(x,y). For such a general case, no conclusions can be drawn in general. Nothing in my answer or previous comments necessarily holds for your edited question. However, if you can state that L(x,y) is bilinear, i.e., linear in y when x is fixed and linear in x when y is fixed, then the methodology I stated above still applies, even though the solution may change.
$endgroup$
– Mark L. Stone
Mar 23 at 19:50
add a comment |
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1 Answer
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$begingroup$
Consider the four optimization problems, which cover the combinations of minimizing or maximizing y, when x is fixed at 0 or 1.
a) min w.r.t. y subject to $0 le y le 1, L(x,y) < 0, x = 0$
b) max w.r.t. y subject to $0 le y le 1, L(x,y) < 0, x = 0$
c) min w.r.t. y subject to $0 le y le 1, L(x,y) < 0, x = 1$
d) max w.r.t. y subject to $0 le y le 1, L(x,y) < 0, x = 1$
Solving these as having non-strict inequality $L(x,y) le 0$, the following solutions are obtained:
a) x = 0, y = 1/3
b) x = 0, y = 1
c) x = 1, y = 0
d) x = 1, y = 2/5.5
Given the linearity of L(x,y) for a fixed x, we can conclude that
when x = 0, then y in the interval (1/3,1] satisfies L(x,y) < 0
when x = 1, then y in the interval [0,2.5/5) satisfies L(x,y) < 0.
Due to the bilinearity of L(x,y), x = 0 and 1 are the extreme cases for x in [0,1] with regard to minimum and maximum possible values of y satisfying L(x,y) < 0.
Noting that 2.5/5 = 0.3626... we can therefore conclude that every value of y in the interval (1/3,2.5/5), and only values of y in that interval, will satisfy L(x,y) < 0 no matter which (i.e., "worst case") value of x in [0,1] is chosen.
answered Mar 23 at 17:42
Mark L. StoneMark L. Stone
1,95058
$endgroup$
$begingroup$
What do you mean when you say linearity? "Given the linearity of L(x,y) for a fixed x" do you mean for a fixed x it is linear in x or y? What exactly is the reasoning behind using edges?
$endgroup$
– physicsP
Mar 23 at 19:28
1
$begingroup$
For a fixed x, L(x,y) is linear in y.
$endgroup$
– Mark L. Stone
Mar 23 at 19:29
$begingroup$
What do you mean by "using edges" in your comment question "What exactly is the reasoning behind using edges?"? Are you asking about my use of the term "extreme cases".? I chose x = 0 and x = 1 as extreme cases, because no intermediate value of x could lead to more extreme bounds on possible y because L(x,y) is linear in y for a fixed x.
$endgroup$
– Mark L. Stone
Mar 23 at 19:44
$begingroup$
You have now changed the question so as to eliminate the specification of particular function L(x,y). For such a general case, no conclusions can be drawn in general. Nothing in my answer or previous comments necessarily holds for your edited question. However, if you can state that L(x,y) is bilinear, i.e., linear in y when x is fixed and linear in x when y is fixed, then the methodology I stated above still applies, even though the solution may change.
$endgroup$
– Mark L. Stone
Mar 23 at 19:50
add a comment |
$begingroup$
Consider the four optimization problems, which cover the combinations of minimizing or maximizing y, when x is fixed at 0 or 1.
a) min w.r.t. y subject to $0 le y le 1, L(x,y) < 0, x = 0$
b) max w.r.t. y subject to $0 le y le 1, L(x,y) < 0, x = 0$
c) min w.r.t. y subject to $0 le y le 1, L(x,y) < 0, x = 1$
d) max w.r.t. y subject to $0 le y le 1, L(x,y) < 0, x = 1$
Solving these as having non-strict inequality $L(x,y) le 0$, the following solutions are obtained:
a) x = 0, y = 1/3
b) x = 0, y = 1
c) x = 1, y = 0
d) x = 1, y = 2/5.5
Given the linearity of L(x,y) for a fixed x, we can conclude that
when x = 0, then y in the interval (1/3,1] satisfies L(x,y) < 0
when x = 1, then y in the interval [0,2.5/5) satisfies L(x,y) < 0.
Due to the bilinearity of L(x,y), x = 0 and 1 are the extreme cases for x in [0,1] with regard to minimum and maximum possible values of y satisfying L(x,y) < 0.
Noting that 2.5/5 = 0.3626... we can therefore conclude that every value of y in the interval (1/3,2.5/5), and only values of y in that interval, will satisfy L(x,y) < 0 no matter which (i.e., "worst case") value of x in [0,1] is chosen.
answered Mar 23 at 17:42
Mark L. StoneMark L. Stone
1,95058
$endgroup$
$begingroup$
What do you mean when you say linearity? "Given the linearity of L(x,y) for a fixed x" do you mean for a fixed x it is linear in x or y? What exactly is the reasoning behind using edges?
$endgroup$
– physicsP
Mar 23 at 19:28
1
$begingroup$
For a fixed x, L(x,y) is linear in y.
$endgroup$
– Mark L. Stone
Mar 23 at 19:29
$begingroup$
What do you mean by "using edges" in your comment question "What exactly is the reasoning behind using edges?"? Are you asking about my use of the term "extreme cases".? I chose x = 0 and x = 1 as extreme cases, because no intermediate value of x could lead to more extreme bounds on possible y because L(x,y) is linear in y for a fixed x.
$endgroup$
– Mark L. Stone
Mar 23 at 19:44
$begingroup$
You have now changed the question so as to eliminate the specification of particular function L(x,y). For such a general case, no conclusions can be drawn in general. Nothing in my answer or previous comments necessarily holds for your edited question. However, if you can state that L(x,y) is bilinear, i.e., linear in y when x is fixed and linear in x when y is fixed, then the methodology I stated above still applies, even though the solution may change.
$endgroup$
– Mark L. Stone
Mar 23 at 19:50
add a comment |
$begingroup$
Consider the four optimization problems, which cover the combinations of minimizing or maximizing y, when x is fixed at 0 or 1.
a) min w.r.t. y subject to $0 le y le 1, L(x,y) < 0, x = 0$
b) max w.r.t. y subject to $0 le y le 1, L(x,y) < 0, x = 0$
c) min w.r.t. y subject to $0 le y le 1, L(x,y) < 0, x = 1$
d) max w.r.t. y subject to $0 le y le 1, L(x,y) < 0, x = 1$
Solving these as having non-strict inequality $L(x,y) le 0$, the following solutions are obtained:
a) x = 0, y = 1/3
b) x = 0, y = 1
c) x = 1, y = 0
d) x = 1, y = 2/5.5
Given the linearity of L(x,y) for a fixed x, we can conclude that
when x = 0, then y in the interval (1/3,1] satisfies L(x,y) < 0
when x = 1, then y in the interval [0,2.5/5) satisfies L(x,y) < 0.
Due to the bilinearity of L(x,y), x = 0 and 1 are the extreme cases for x in [0,1] with regard to minimum and maximum possible values of y satisfying L(x,y) < 0.
Noting that 2.5/5 = 0.3626... we can therefore conclude that every value of y in the interval (1/3,2.5/5), and only values of y in that interval, will satisfy L(x,y) < 0 no matter which (i.e., "worst case") value of x in [0,1] is chosen.
answered Mar 23 at 17:42
Mark L. StoneMark L. Stone
1,95058
$endgroup$
Consider the four optimization problems, which cover the combinations of minimizing or maximizing y, when x is fixed at 0 or 1.
a) min w.r.t. y subject to $0 le y le 1, L(x,y) < 0, x = 0$
b) max w.r.t. y subject to $0 le y le 1, L(x,y) < 0, x = 0$
c) min w.r.t. y subject to $0 le y le 1, L(x,y) < 0, x = 1$
d) max w.r.t. y subject to $0 le y le 1, L(x,y) < 0, x = 1$
Solving these as having non-strict inequality $L(x,y) le 0$, the following solutions are obtained:
a) x = 0, y = 1/3
b) x = 0, y = 1
c) x = 1, y = 0
d) x = 1, y = 2/5.5
Given the linearity of L(x,y) for a fixed x, we can conclude that
when x = 0, then y in the interval (1/3,1] satisfies L(x,y) < 0
when x = 1, then y in the interval [0,2.5/5) satisfies L(x,y) < 0.
Due to the bilinearity of L(x,y), x = 0 and 1 are the extreme cases for x in [0,1] with regard to minimum and maximum possible values of y satisfying L(x,y) < 0.
Noting that 2.5/5 = 0.3626... we can therefore conclude that every value of y in the interval (1/3,2.5/5), and only values of y in that interval, will satisfy L(x,y) < 0 no matter which (i.e., "worst case") value of x in [0,1] is chosen.
answered Mar 23 at 17:42
Mark L. StoneMark L. Stone
1,95058
edited Mar 23 at 17:48
answered Mar 23 at 17:42
Mark L. StoneMark L. Stone
1,95058
answered Mar 23 at 17:42
Mark L. StoneMark L. Stone
1,95058
answered Mar 23 at 17:42
Mark L. StoneMark L. Stone
1,95058
1,95058
$begingroup$
What do you mean when you say linearity? "Given the linearity of L(x,y) for a fixed x" do you mean for a fixed x it is linear in x or y? What exactly is the reasoning behind using edges?
$endgroup$
– physicsP
Mar 23 at 19:28
1
$begingroup$
For a fixed x, L(x,y) is linear in y.
$endgroup$
– Mark L. Stone
Mar 23 at 19:29
$begingroup$
What do you mean by "using edges" in your comment question "What exactly is the reasoning behind using edges?"? Are you asking about my use of the term "extreme cases".? I chose x = 0 and x = 1 as extreme cases, because no intermediate value of x could lead to more extreme bounds on possible y because L(x,y) is linear in y for a fixed x.
$endgroup$
– Mark L. Stone
Mar 23 at 19:44
$begingroup$
You have now changed the question so as to eliminate the specification of particular function L(x,y). For such a general case, no conclusions can be drawn in general. Nothing in my answer or previous comments necessarily holds for your edited question. However, if you can state that L(x,y) is bilinear, i.e., linear in y when x is fixed and linear in x when y is fixed, then the methodology I stated above still applies, even though the solution may change.
$endgroup$
– Mark L. Stone
Mar 23 at 19:50
add a comment |
$begingroup$
What do you mean when you say linearity? "Given the linearity of L(x,y) for a fixed x" do you mean for a fixed x it is linear in x or y? What exactly is the reasoning behind using edges?
$endgroup$
– physicsP
Mar 23 at 19:28
1
$begingroup$
For a fixed x, L(x,y) is linear in y.
$endgroup$
– Mark L. Stone
Mar 23 at 19:29
$begingroup$
What do you mean by "using edges" in your comment question "What exactly is the reasoning behind using edges?"? Are you asking about my use of the term "extreme cases".? I chose x = 0 and x = 1 as extreme cases, because no intermediate value of x could lead to more extreme bounds on possible y because L(x,y) is linear in y for a fixed x.
$endgroup$
– Mark L. Stone
Mar 23 at 19:44
$begingroup$
You have now changed the question so as to eliminate the specification of particular function L(x,y). For such a general case, no conclusions can be drawn in general. Nothing in my answer or previous comments necessarily holds for your edited question. However, if you can state that L(x,y) is bilinear, i.e., linear in y when x is fixed and linear in x when y is fixed, then the methodology I stated above still applies, even though the solution may change.
$endgroup$
– Mark L. Stone
Mar 23 at 19:50
$begingroup$
What do you mean when you say linearity? "Given the linearity of L(x,y) for a fixed x" do you mean for a fixed x it is linear in x or y? What exactly is the reasoning behind using edges?
$endgroup$
– physicsP
Mar 23 at 19:28
$begingroup$
What do you mean when you say linearity? "Given the linearity of L(x,y) for a fixed x" do you mean for a fixed x it is linear in x or y? What exactly is the reasoning behind using edges?
$endgroup$
– physicsP
Mar 23 at 19:28
1
1
$begingroup$
For a fixed x, L(x,y) is linear in y.
$endgroup$
– Mark L. Stone
Mar 23 at 19:29
$begingroup$
For a fixed x, L(x,y) is linear in y.
$endgroup$
– Mark L. Stone
Mar 23 at 19:29
$begingroup$
What do you mean by "using edges" in your comment question "What exactly is the reasoning behind using edges?"? Are you asking about my use of the term "extreme cases".? I chose x = 0 and x = 1 as extreme cases, because no intermediate value of x could lead to more extreme bounds on possible y because L(x,y) is linear in y for a fixed x.
$endgroup$
– Mark L. Stone
Mar 23 at 19:44
$begingroup$
What do you mean by "using edges" in your comment question "What exactly is the reasoning behind using edges?"? Are you asking about my use of the term "extreme cases".? I chose x = 0 and x = 1 as extreme cases, because no intermediate value of x could lead to more extreme bounds on possible y because L(x,y) is linear in y for a fixed x.
$endgroup$
– Mark L. Stone
Mar 23 at 19:44
$begingroup$
You have now changed the question so as to eliminate the specification of particular function L(x,y). For such a general case, no conclusions can be drawn in general. Nothing in my answer or previous comments necessarily holds for your edited question. However, if you can state that L(x,y) is bilinear, i.e., linear in y when x is fixed and linear in x when y is fixed, then the methodology I stated above still applies, even though the solution may change.
$endgroup$
– Mark L. Stone
Mar 23 at 19:50
$begingroup$
You have now changed the question so as to eliminate the specification of particular function L(x,y). For such a general case, no conclusions can be drawn in general. Nothing in my answer or previous comments necessarily holds for your edited question. However, if you can state that L(x,y) is bilinear, i.e., linear in y when x is fixed and linear in x when y is fixed, then the methodology I stated above still applies, even though the solution may change.
$endgroup$
– Mark L. Stone
Mar 23 at 19:50
add a comment |
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Required, but never shown
Required, but never shown
