# Finding maximum value

1. Mar 12, 2013

### Saitama

1. The problem statement, all variables and given/known data
Find the largest possible value of
$$\sqrt{(x-20)(y-x)}+\sqrt{(140-y)(20-x)}+\sqrt{(x-y)(y-140)}$$
subject to $-40≤x≤100$ and $-20≤y≤200$.

2. Relevant equations

3. The attempt at a solution
Factoring out $\sqrt{(x-20)(y-x)(140-y)}$ the given expression can be written as
$$\sqrt{(x-20)(y-x)(140-y)}\left(\frac{1}{\sqrt{140-y}}+\frac{1}{\sqrt{x-y}}+\frac{1}{\sqrt{x-20}}\right)$$
From the above equation, $x≠y$, $x>20$, $x>y$ and $y<140$.
Hence, $20≤x≤100$ and $-20≤y≤100$.
But now I am stuck here. I have no clue about how should I proceed further. :(

Any help is appreciated. Thanks!

2. Mar 12, 2013

### jingu

hi

I f we take cases for inequality between x and y ,none of them satisfy.

3. Mar 12, 2013

### Saitama

Not sure what you mean, can you elaborate a bit more?

4. Mar 12, 2013

### sankalpmittal

Have you tried applying "Arithematic Mean ≥ Geometric mean" concept on each term? I can see a neat and clear approach from there.

For example : The first term :

√{(x−20)(y−x)} ≤ {(x-20)+(y-x)}/2

5. Mar 12, 2013

### Saitama

How does that inequality help here? Applying that on the first term, I end up with
$$\sqrt{(x-20)(y-x)} \leq \frac{y-20}{2}$$

How to proceed from here?

6. Mar 12, 2013

### sankalpmittal

Do you not notice that the maximum value of the first term is (y-20)/2 ? Keep on applying this for other two remaining terms and find their maximum value also.

7. Mar 12, 2013

### Saitama

Okay, so you mean y=200?

8. Mar 12, 2013

### sankalpmittal

No hurry. Leave (y-20)/2 as it is for now. Find the maximum value for other terms using A.M.≥G.M. inequality in terms of x and y. What do you get ?

9. Mar 12, 2013

### Saitama

For the second term it is $\leq \frac{20-y-x}{2}$ and for the third term it is $\leq \frac{x-140}{2}$.

10. Mar 12, 2013

### sankalpmittal

For second term it should be (160-y-x)/2.. Now you know the maximum value for each term. Now add each term up. Obviously, adding each term will evaluate maximum possible value.

11. Mar 12, 2013

### Saitama

Oops, yes, it is 160.

Not sure but would you set x=100 and y=200 for the maximum value?

12. Mar 12, 2013

### sankalpmittal

That is not going to work. I am getting negative value for that. Check. Again, by using several values for y and x you get negative values. Either try plugging them in original expression or modified. Even worse, 140>y and y>140, both should be possible in original expression to avoid complex roots. How in the world do you think this will be possible?

Take several ordered pairs of (x,y) and plug them in modified expression. You will get a negative values for majority.

So instead simply add the maximum values you obtained for each term using inequality. You will see that x and y automatically cancel.

13. Mar 12, 2013

### Saitama

:uhh:
If I add the expressions for maximum values, it turns out to be zero. :(

14. Mar 12, 2013

### pcm

That is not going to work.What you say is maximum is true only for a=b,if you apply AM GM inequality on a,b.Doing so for 1st 2 expressions give x=140,y=260.

*I don't know the answer*

15. Mar 12, 2013

### sankalpmittal

Edit: Ok, do not add it up. Probably plug (x,y) in modified expression to seek maximum magnitude of expression.

I again post this wordings :

Even worse, 140>y and y>140, both should be possible in original expression to avoid complex roots. How in the world do you think this will be possible?

That is only possible when y=x. Also just analyze middle term in original expression.

Analyzing the original expression further, to avoid complex roots :

Case 1:

x>20
y>x => y>20

y<140
x<20

x>y
y>140

Contradictions. You get complex roots here.

Case 2:

x<20 and y<x => y<20
y>140 and x>20
y>x and y<140

Again contradictions..

To avoid contradictions, I can see that y=x is the only way.

@PCM:
y=260 is out of domain.

Last edited: Mar 12, 2013
16. Mar 12, 2013

### pcm

(10,10)

17. Mar 12, 2013

### sankalpmittal

(-20,-20).. That is better. Sorry for messing this up and not analyzing the middle term carefully.

That inequality was probably hazy thing to apply here. To avoid complex roots, x=y has to be possible.

Also see my previous post.

Last edited: Mar 12, 2013
18. Mar 12, 2013

### pcm

yes,x=y is only possible.
By graphing all lines and checking possible values for x,
i get x between -20 and 20.
And maximum is indeed for x=-20.

19. Mar 12, 2013

### Saitama

If I use y=x, the original expression reduces $\sqrt{(140-x)(20-x)}$, can I use Calculus to find its maximum value?

20. Mar 13, 2013

### jingu

I got it

When we put x=y, and x=y=-20, we get the maximum value as 80.

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