# Finding maximum value

1. Jan 28, 2014

### utkarshakash

1. The problem statement, all variables and given/known data
For a,b belongs to R the maximum value of $(a-1)(b-1)+(1-\sqrt{1-a^2})(1-\sqrt{1-b^2})$

2. Relevant equations

3. 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.

Last edited: Jan 28, 2014
2. Jan 28, 2014

### dirk_mec1

Lagrangre? Your transformation is also not correct.

3. Jan 28, 2014

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.

4. Jan 28, 2014

### dirk_mec1

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

What does this mean for your equation?

5. Jan 28, 2014

### Ray Vickson

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.)

6. Jan 28, 2014

### haruspex

No, Lagrange.

7. Jan 28, 2014

### haruspex

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

8. Jan 28, 2014

### haruspex

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. Jan 28, 2014

### Ray Vickson

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 $b$.

10. Jan 28, 2014

### haruspex

The substitution of trig functions handles that automatically here, no?

11. Jan 28, 2014

### Ray Vickson

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. Jan 28, 2014

### Ray Vickson

I don't see why it is easier to substitute $\theta = 0 \text{ or } \pi$ instead of $a = -1 \text{ or } 1$.

13. Jan 28, 2014

### Ray Vickson

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. Jan 28, 2014

### haruspex

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. Jan 28, 2014

### Ray Vickson

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.

16. Jan 29, 2014

### utkarshakash

Your method is pretty much easier than others but I am not able to realise why the maximum should occur when a=b

17. Jan 29, 2014

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. Jan 29, 2014

### Ray Vickson

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. Jan 29, 2014

### haruspex

Wrong. If f and g reach their maxima for different values of x then max{f+g} < max{f}+max{g}.

20. Jan 29, 2014

### utkarshakash

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