# Exam : find the minimum value of this function and constrant

1. Oct 9, 2011

### ~Scott~

1. The problem statement, all variables and given/known data
Given
$$x+y = 6\sqrt2$$
$$a + b = 6\sqrt 2$$
Find minimum value of $$\sqrt{x^2 + a^2} + \sqrt{y^2+b^2}$$

2. The attempt at a solution
I had plotted the $$f(x,y)$$ with constant $$a ,b$$
And that's all. I don't realize how to do this.

Any comment will be appreciated

Last edited: Oct 9, 2011
2. Oct 9, 2011

### verty

I don't know where on earth you found this question, but one can find the minimum value. The key step is to notice that you can simplify the total derivative considerably using the relation between x and y. That, together with...

But I've probably said enough. I figure this is a take-home exam or something.

Last edited: Oct 9, 2011
3. Oct 9, 2011

### SammyS

Staff Emeritus
If this is a take home exam, we really shouldn't help you.

4. Oct 9, 2011

### Dick

It's a Lagrange multiplier problem with two constraints. Try that technique.

5. Oct 10, 2011

### ~Scott~

Let me introduce myself. Sorry to take pity on me.
I graduated from Physics and now studying in computer science.

Of course, I studied Lagrange multiplier in classical mechanics.

I always spend my free time to enjoy mathematics and computing.
I found this problem from math.or.th age not over 15.
And I never expect that I have to use Lagrange multiplier.

You can check my post about GPU/CUDA topics that is my first post.
And now it's finish without any reply from this forum.

And second post of mine is about gaussian gun. Again no one reply.

The third one is this about lagrange multiplier.

Guess what.
Your people totally look down on me just because you known how to solve it
What about two questions of mine ?

I know your situation that many lazy students exploit this forum for their benefit.
It's ok, but please count me out.

If anyone would like me to identify myself as a master degree in computer science. Just sent me your PM and I will show it to you.

Thanks everyone who show me the topics of Lagrange multiplier.

6. Oct 10, 2011

### Ray Vickson

There is another way: substitute y = c-x into f(x,y), where c = 6*sqrt(2). The result F(x) = f(x,c-x) is a univariate function, and can be minimized by setting its derivative to zero; in fact, we know the function F is strictly convex, so any stationary point is a global minimum.

Finding the derivative is not too hard, but solving the equation F'(x) = 0 for x is a bit nasty. (Here, I am assuming that a and b are not variables, but are just some constants that satisfy a+b=c.) It turns out that the solution is just x = a (when the relationship between a and b is exploited). I took the easy way out, and let Maple 11 solve the equations for me; doing the problem by hand would be difficult.

RGV

7. Oct 11, 2011

### ~Scott~

Thank you RGV.
I will check it.