# Inequalities

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#2

n = ...

Then replace the variables to make an inequality? What are the min/max values.

Or start with the two inequalities as they're reasonably straightforward.

Last edited by mqb2766; 1 month ago

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#4

... <= n <= ...

Then argue about the min/max value of the expressions if you don't simplify them first.

Last edited by mqb2766; 1 month ago

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Is it wrong if I take - 4pi and 12pi as min and max values for the above equation for all alpha?

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#6

(Original post by

Is it wrong if I take - 4pi and 12pi as min and max values for the above equation for all alpha?

**Nimantha**)Is it wrong if I take - 4pi and 12pi as min and max values for the above equation for all alpha?

Note you also have x to bound.

Last edited by mqb2766; 1 month ago

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#8

n = ...

To find the actual expression in x, alpha you want to find the min/max value of which satisfies the constraints. Maybe try that for starters?

In doing so, you may have to consider x=0 division, but it should give insight.

Last edited by mqb2766; 1 month ago

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(Original post by

I would have written

n = ...

To find the actual expression in x, alpha you want to find the min/max value of which satisfies the constraints. Maybe try that for starters?

**mqb2766**)I would have written

n = ...

To find the actual expression in x, alpha you want to find the min/max value of which satisfies the constraints. Maybe try that for starters?

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#10

(Original post by

Ok thanks...

**Nimantha**)Ok thanks...

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#11

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The original question was this, and I got stuck in the second part. And I think there is a problem in this question as it doesn't give the same number of solutions for all alpha in the given range

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#14

(Original post by

But it doesn't work for all alpha in the range like pi/6

**3672N**)But it doesn't work for all alpha in the range like pi/6

Edit, when looking at the original question, alpha is fixed (determined by a and b (h?)), not a free variable, which obviously changes the problem. Is the final inequality constraint correct

2pi - alpha <= x <= 2pi+alpha

So for instance if alpha=0, x=2pi and there cannot be a stationary point as the range of x is a point? Or do you want to take the range extreme value as the max value?

Note there appears to be some ambiguity in the question, depending on the sign of B the location of the maximum could be 3pi/2 rather than pi/2. I take it you've accounted for that?

Last edited by mqb2766; 1 month ago

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(Original post by

The way you asked it was that alpha was a slack variable that just had to satisfy the inequality constant. So as long as there exists one alpha in that range, the solution is fine.

Edit, when looking at the original question, alpha is fixed (determined by a and b (h?)), not a free variable, which obviously changes the problem. Is the final inequality constraint correct

2pi - alpha <= x <= 2pi+alpha

So for instance if alpha=0, x=2pi and there cannot be a stationary point as the range of x is a point? Or do you want to take the range extreme value as the max value?

Note there appears to be some ambiguity in the question, depending on the sign of B the location of the maximum could be 3pi/2 rather than pi/2. I take it you've accounted for that?

**mqb2766**)The way you asked it was that alpha was a slack variable that just had to satisfy the inequality constant. So as long as there exists one alpha in that range, the solution is fine.

Edit, when looking at the original question, alpha is fixed (determined by a and b (h?)), not a free variable, which obviously changes the problem. Is the final inequality constraint correct

2pi - alpha <= x <= 2pi+alpha

So for instance if alpha=0, x=2pi and there cannot be a stationary point as the range of x is a point? Or do you want to take the range extreme value as the max value?

Note there appears to be some ambiguity in the question, depending on the sign of B the location of the maximum could be 3pi/2 rather than pi/2. I take it you've accounted for that?

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#16

(Original post by

h and b are not the same, and the final ineqality is correct. B=1/2sqrt((a-b)^2+4h^2), which we can take as positive, so the maximum must be at pi/2

**3672N**)h and b are not the same, and the final ineqality is correct. B=1/2sqrt((a-b)^2+4h^2), which we can take as positive, so the maximum must be at pi/2

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**mqb2766**)

The way you asked it was that alpha was a slack variable that just had to satisfy the inequality constant. So as long as there exists one alpha in that range, the solution is fine.

Edit, when looking at the original question, alpha is fixed (determined by a and b (h?)), not a free variable, which obviously changes the problem. Is the final inequality constraint correct

2pi - alpha <= x <= 2pi+alpha

So for instance if alpha=0, x=2pi and there cannot be a stationary point as the range of x is a point? Or do you want to take the range extreme value as the max value?

Note there appears to be some ambiguity in the question, depending on the sign of B the location of the maximum could be 3pi/2 rather than pi/2. I take it you've accounted for that?

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#18

(Original post by

I don't undestand what did you mean by the solution is fine as long as there exist one one alpha, cause there can't be a general solution for n as, the number of solutions for n changes as alpha increases or decreses by pi

**3672N**)I don't undestand what did you mean by the solution is fine as long as there exist one one alpha, cause there can't be a general solution for n as, the number of solutions for n changes as alpha increases or decreses by pi

Last edited by mqb2766; 1 month ago

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(Original post by

Thats how I was interpreting the problem last night. Seeing the full question, its obviously incorrect as alpha is fixed by the initial transformation.

**mqb2766**)Thats how I was interpreting the problem last night. Seeing the full question, its obviously incorrect as alpha is fixed by the initial transformation.

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(Original post by

what does the final inequality represent?

**mqb2766**)what does the final inequality represent?

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