Maximize Value: Solving for Largest Possible Value Using Factoring Method

[Saitama]

Homework Statement

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 $-20≤y≤200$.

Homework Equations

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!

 

[jingu]

hi

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

 

[Saitama]

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

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

 

[sankalpmittal]

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

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

 

[Saitama]

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

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?

 

[sankalpmittal]

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?

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.

 

[Saitama]

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.

Okay, so you mean y=200?

 

[sankalpmittal]

Okay, so you mean y=200?

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 ?

 

[Saitama]

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 ?

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

 

[sankalpmittal]

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

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.

 

[Saitama]

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.

Oops, yes, it is 160.

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

 

[sankalpmittal]

Oops, yes, it is 160.

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

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.

 

[Saitama]

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

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

 

[pcm]

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

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*

 

[sankalpmittal]

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

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:

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*

y=260 is out of domain.

 

[pcm]

(10,10)

 

[sankalpmittal]

(10,10)

(-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.

 

[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.

 

[Saitama]

(-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.

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

 

[jingu]

I got it

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

 

[ehild]

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

You get the extreme, but it is not maximum (why?)

ehild

 

[Saitama]

You get the extreme, but it is not maximum (why?)

ehild

Oh yes, it is not maximum. It was obvious once I analysed the function. Thanks to everyone for the help. :-)