# Finding maximum value

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## Homework Statement

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

## 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 $y=\sqrt{1-x^2}$. Now by inspection the original function transforms to $(x_1 - 1 )(x_2 - 1) + (1-y_1)(1-y_2)$.

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

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Lagrangre? Your transformation is also not correct.

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-$\sqrt{1-a^{2}}$)(1-$\sqrt{1-b^{2}}$) 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-$\sqrt{1-a^{2}}$)(1-$\sqrt{1-b^{2}}$) by first maximising (1-$\sqrt{1-a^{2}}$) with the help of differentiation and doing the same for (1-$\sqrt{1-b^{2}}$).

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.

Hint: do you think that for the maximum that a = b ? :D

What does this mean for your equation?

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

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

## 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 $y=\sqrt{1-x^2}$. Now by inspection the original function transforms to $(x_1 - 1 )(x_2 - 1) + (1-y_1)(1-y_2)$.

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.)

haruspex
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Lagrangre?
No, Lagrange.

haruspex
Homework Helper
Gold Member
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-$\sqrt{1-a^{2}}$)(1-$\sqrt{1-b^{2}}$) 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-$\sqrt{1-a^{2}}$)(1-$\sqrt{1-b^{2}}$) by first maximising (1-$\sqrt{1-a^{2}}$) with the help of differentiation and doing the same for (1-$\sqrt{1-b^{2}}$).

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.
I fail to see how maximising the terms independently will work. Are you confusing [-1, 1] with {-1, 1}?

haruspex
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Gold Member
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.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

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

## 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 $y=\sqrt{1-x^2}$. Now by inspection the original function transforms to $(x_1 - 1 )(x_2 - 1) + (1-y_1)(1-y_2)$.

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:
$$\frac{\partial f}{\partial a} = \begin{cases} 0 &\text{ if } -1 < a < 1\\ \leq 0 &\text{ if } a = -1 \\ \geq 0 &\text{ if } a = +1 \end{cases}$$
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##.

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?

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.

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##.

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.

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.

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
$$g(\theta,\phi) = (\cos(\theta)-1)(\cos(\phi)-1) + (1-\sin(\theta))(1-\sin(\phi))$$
respresents the original function
$$f(a,b) = (a-1)(b-1) + (1-\sqrt{1-a^2})(1-\sqrt{1-b^2})$$
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.

Gold Member
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

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.

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.

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}.

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So how should I start? You already said that trig substitution does not work.

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)).

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.