Finding maximum value

  • #1
utkarshakash
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Homework Statement


For a,b belongs to R the maximum value of [itex](a-1)(b-1)+(1-\sqrt{1-a^2})(1-\sqrt{1-b^2})[/itex]


Homework Equations



The Attempt at a Solution


I'm really clueless regarding how to start. So I started putting some arbitrary values but they were of no help. I can't even differentiate it as it is not a function in one variable. But it is clear that whatever value a and b attains, the following inequalities must be satisfied.

1-a^2>0
1-b^2>0

This means that a and b must lie between -1 and 1. So far I could shorten the range of a and b. I also remember that I posted a very similar question earlier and someone suggested me to use geometric methods rather than solving it algebraically. OK, so I assume a function [itex]y=\sqrt{1-x^2}[/itex]. Now by inspection the original function transforms to [itex](x_1 - 1 )(x_2 - 1) + (1-y_1)(1-y_2)[/itex].

But I really don't know how this helps me.
 
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Answers and Replies

  • #2
761
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Lagrangre? Your transformation is also not correct.
 
  • #3
113
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You are proceeding in the right way! You have found out the range of 'a' and 'b'. Now this is just like finding the maximum of

(a-1)(b-1) + (1-[itex]\sqrt{1-a^{2}}[/itex])(1-[itex]\sqrt{1-b^{2}}[/itex]) when a,bε[-1,1]

For that maximize the first expression i.e, (a-1)(b-1) by substituting the appropriate value from the set[-1,1]; and maximize the second by i.e, (1-[itex]\sqrt{1-a^{2}}[/itex])(1-[itex]\sqrt{1-b^{2}}[/itex]) by first maximising (1-[itex]\sqrt{1-a^{2}}[/itex]) with the help of differentiation and doing the same for (1-[itex]\sqrt{1-b^{2}}[/itex]).

Summarizing, all I want to say is, separately maximize each of the component by substituting numbers from [-1,1].

Hope you get it. Best of luck!

Regards.
ADI
 
  • #4
761
13
Hint: do you think that for the maximum that a = b ? :D

What does this mean for your equation?
 
  • #5
Ray Vickson
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Homework Statement


For a,b belongs to R the maximum value of [itex](a-1)(b-1)+(1-\sqrt{1-a^2})(1-\sqrt{1-b^2})[/itex]


Homework Equations



The Attempt at a Solution


I'm really clueless regarding how to start. So I started putting some arbitrary values but they were of no help. I can't even differentiate it as it is not a function in one variable. But it is clear that whatever value a and b attains, the following inequalities must be satisfied.

1-a^2>0
1-b^2>0

This means that a and b must lie between -1 and 1. So far I could shorten the range of a and b. I also remember that I posted a very similar question earlier and someone suggested me to use geometric methods rather than solving it algebraically. OK, so I assume a function [itex]y=\sqrt{1-x^2}[/itex]. Now by inspection the original function transforms to [itex](x_1 - 1 )(x_2 - 1) + (1-y_1)(1-y_2)[/itex].

But I really don't know how this helps me.
You should not claim 1-a^2 > 0 and 1-b^2 > 0; you just need 1 - a^2 >= 0 and 1-b^2 >= 0. That makes all the difference in the world: with strict inequalities you have an open set of feasible (a,b) values, and so there might not be a maximum at all. With non-strict inequalities you get a closed set, so are guaranteed the existence of a maximum (and minimum). Also, because you have inequalilty constraints on the variables you cannot be sure that differentiation will lead to a proper solution. (Hint: in this case it does not.)
 
  • #7
haruspex
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You are proceeding in the right way! You have found out the range of 'a' and 'b'. Now this is just like finding the maximum of

(a-1)(b-1) + (1-[itex]\sqrt{1-a^{2}}[/itex])(1-[itex]\sqrt{1-b^{2}}[/itex]) when a,bε[-1,1]

For that maximize the first expression i.e, (a-1)(b-1) by substituting the appropriate value from the set[-1,1]; and maximize the second by i.e, (1-[itex]\sqrt{1-a^{2}}[/itex])(1-[itex]\sqrt{1-b^{2}}[/itex]) by first maximising (1-[itex]\sqrt{1-a^{2}}[/itex]) with the help of differentiation and doing the same for (1-[itex]\sqrt{1-b^{2}}[/itex]).

Summarizing, all I want to say is, separately maximize each of the component by substituting numbers from [-1,1].

Hope you get it. Best of luck!

Regards.
ADI
I fail to see how maximising the terms independently will work. Are you confusing [-1, 1] with {-1, 1}?
 
  • #8
haruspex
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My immediate thought is to substitute trig functions. Should make the calculus a bit easier, but it doesn't help as much as I'd hoped.
 
  • #9
Ray Vickson
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Homework Statement


For a,b belongs to R the maximum value of [itex](a-1)(b-1)+(1-\sqrt{1-a^2})(1-\sqrt{1-b^2})[/itex]


Homework Equations



The Attempt at a Solution


I'm really clueless regarding how to start. So I started putting some arbitrary values but they were of no help. I can't even differentiate it as it is not a function in one variable. But it is clear that whatever value a and b attains, the following inequalities must be satisfied.

1-a^2>0
1-b^2>0

This means that a and b must lie between -1 and 1. So far I could shorten the range of a and b. I also remember that I posted a very similar question earlier and someone suggested me to use geometric methods rather than solving it algebraically. OK, so I assume a function [itex]y=\sqrt{1-x^2}[/itex]. Now by inspection the original function transforms to [itex](x_1 - 1 )(x_2 - 1) + (1-y_1)(1-y_2)[/itex].

But I really don't know how this helps me.
If you know about derivatives, the necessary conditions for a maximum of ##f(a,b)## = your function are:
[tex] \frac{\partial f}{\partial a} = \begin{cases}
0 &\text{ if } -1 < a < 1\\
\leq 0 &\text{ if } a = -1 \\
\geq 0 &\text{ if } a = +1
\end{cases}[/tex]
with the same type of conditions for ##b##. This is not very useful in itself, but it does act as a reminder to look not only at interior points, but also at the endpoints of the intervals for ##a## and ##b##.
 
  • #10
haruspex
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it does act as a reminder to look not only at interior points, but also at the endpoints of the intervals
The substitution of trig functions handles that automatically here, no?
 
  • #11
Ray Vickson
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The substitution of trig functions handles that automatically here, no?
I don't see why it is easier to substitute ##\theta = \pi \text{ or } 0## instead of ##a = -1 \text{ or } 1##. Admittedly, writing ##a = \cos(\theta)## does restrict ##a## to the desired interval, but the endpoints still need checking manually.
 
  • #12
Ray Vickson
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The substitution of trig functions handles that automatically here, no?
I don't see why it is easier to substitute ##\theta = 0 \text{ or } \pi## instead of ##a = -1 \text{ or } 1##.
 
  • #13
Ray Vickson
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The substitution of trig functions handles that automatically here, no?
I don't see why it is easier to substitute ##\theta = 0 \text{ or } \pi## instead of ##a = -1 \text{ or } 1##. Admittedly, the substitution ##a =\cos(\theta)## does automatically restrict ##a## to the correct interval, but the endpoints still need to be checked manually.
 
  • #14
haruspex
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the substitution ##a =\cos(\theta)## does automatically restrict ##a## to the correct interval, but the endpoints still need to be checked manually.
Are you sure? That is not clear to me. If the max is at a = 1 then as theta passes through 0 the function will reach the max then fall off again, making it a local maximum.
 
  • #15
Ray Vickson
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Are you sure? That is not clear to me. If the max is at a = 1 then as theta passes through 0 the function will reach the max then fall off again, making it a local maximum.
The function
[tex] g(\theta,\phi) = (\cos(\theta)-1)(\cos(\phi)-1) + (1-\sin(\theta))(1-\sin(\phi))[/tex]
respresents the original function
[tex] f(a,b) = (a-1)(b-1) + (1-\sqrt{1-a^2})(1-\sqrt{1-b^2})[/tex]
only for ##0 \leq \theta, \phi \leq \pi##. If ##\theta## or ##\phi## are a bit less than 0 or a bit more than ##\pi##, the "sin" parts go negative and so no longer represent the square roots in f.

In this problem the partial derivatives of f are not both zero at the maximum, and the partial derivatives of g are also not both zero at the maximum. I don't want to say more now, because I don't want to give away the solution.
 
  • #16
utkarshakash
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Hint: do you think that for the maximum that a = b ? :D

What does this mean for your equation?
Your method is pretty much easier than others but I am not able to realise why the maximum should occur when a=b
 
  • #17
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I fail to see how maximising the terms independently will work. Are you confusing [-1, 1] with {-1, 1}?
No, I am not. For finding the maximum value of two monotonous functions f(x) + g(x) we find the max value of each and then add them together, right? Since the values of a and b are confined to [-1,1], substituting the two extreme values i.e, -1 or 1 has a possibility of getting us the max value.
 
  • #18
Ray Vickson
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Your method is pretty much easier than others but I am not able to realise why the maximum should occur when a=b
The suggestion to take a = b is based on the symmetry of f(a,b). However, it is dangerous: just having symmetry does not imply that a = b at the solution of an optimization problem. Symmetry just implies that if (a,b) = (u,v) is a solution, then (a,b) = (v,u) is also a solution. This actually happens in this problem for the minimum of f! However, I won't say what happens for the maximum.
 
  • #19
haruspex
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No, I am not. For finding the maximum value of two monotonous functions f(x) + g(x) we find the max value of each and then add them together, right?
Wrong. If f and g reach their maxima for different values of x then max{f+g} < max{f}+max{g}.
 
  • #20
utkarshakash
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So how should I start? You already said that trig substitution does not work.
 
  • #21
haruspex
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So how should I start? You already said that trig substitution does not work.
Who said it doesn't work? I said it didn't make it really easy; Ray thinks it's not valid (or requires great care, maybe).
In general, to find the max of a differentiable function f=f(x,y), where (x,y) lies in some rectangle, involves considering all combinations of {x : either ∂f/∂x = 0 or x is at an extreme value} and {y : either ∂f/∂y = 0 or y is at an extreme value}, 9 cases in all. (The colon indicates "such that".)
If, in the present case, we substitute a = cos(θ), b = cos(ϕ) then f = (1-cos(θ))(1-cos(ϕ))+(1-sin(θ))(1-sin(ϕ)), where θ and ϕ are in [0, π]. (Note that this meets the requirement that √ is to be taken as meaning the non-negative root.) We need to look at how ∂f/∂x relates to ∂f/∂θ etc:
∂f/∂θ = -∂f/∂x sin(θ)
It follows that the condition ∂f/∂θ = 0 is equivalent to "{∂f/∂x = 0} or {sin(θ)=0}". But sin(θ)=0 corresponds to the endpoints of the range, so {x : ∂f/∂θ = 0} = {x : either ∂f/∂x = 0 or x is at an extreme value}, exactly what we wanted. So we have reduced the problem to finding simultaneous solutions of ∂f/∂θ = 0 and ∂f/∂ϕ = 0, then checking to see which values actually give the maximum.
So, go ahead and differentiate and see where you get to. A trick you might find useful is cos(θ)-sin(θ) = √2(cos(θ+π/4)).
 
  • #22
Ray Vickson
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Who said it doesn't work? I said it didn't make it really easy; Ray thinks it's not valid (or requires great care, maybe).
In general, to find the max of a differentiable function f=f(x,y), where (x,y) lies in some rectangle, involves considering all combinations of {x : either ∂f/∂x = 0 or x is at an extreme value} and {y : either ∂f/∂y = 0 or y is at an extreme value}, 9 cases in all. (The colon indicates "such that".)
If, in the present case, we substitute a = cos(θ), b = cos(ϕ) then f = (1-cos(θ))(1-cos(ϕ))+(1-sin(θ))(1-sin(ϕ)), where θ and ϕ are in [0, π]. (Note that this meets the requirement that √ is to be taken as meaning the non-negative root.) We need to look at how ∂f/∂x relates to ∂f/∂θ etc:
∂f/∂θ = -∂f/∂x sin(θ)
It follows that the condition ∂f/∂θ = 0 is equivalent to "{∂f/∂x = 0} or {sin(θ)=0}". But sin(θ)=0 corresponds to the endpoints of the range, so {x : ∂f/∂θ = 0} = {x : either ∂f/∂x = 0 or x is at an extreme value}, exactly what we wanted. So we have reduced the problem to finding simultaneous solutions of ∂f/∂θ = 0 and ∂f/∂ϕ = 0, then checking to see which values actually give the maximum.
So, go ahead and differentiate and see where you get to. A trick you might find useful is cos(θ)-sin(θ) = √2(cos(θ+π/4)).
I did not say it was invalid; I just said it required care, and that I could not see how it made anything easier. Admittedly, the partial derivatives are a lot easier to write in the trig case, but solving for stationary points (if any) is not really easier.
 

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