Algebraic/Geometric Manipulations

Geometry Level 5

$\displaystyle{{ x }+{ y }={ \left( \cfrac { \sqrt { x } }{ \sqrt { y } } +\cfrac { \sqrt { y } }{ \sqrt { x } } \right) }^{ 2 }}$

$x$ and $y$ are positive real numbers satisfying the above equation.

Find Maximum value of $\displaystyle{\cfrac { 24\sqrt { x } +7\sqrt { y } +31\sqrt { xy } }{ \sqrt { xy } } }$

Credits :

Megh choksi

The answer is 56.

2 solutions

Andrea Palma
Mar 20, 2015

For $x, y \in \mathbb R^{+}$ the condition on the variables is equivalent to

$\frac{x+y}{xy} = 1$

Note that

$\displaystyle{\cfrac { 24\sqrt { x } +7\sqrt { y } +31\sqrt { xy } }{ \sqrt { xy } } } = \displaystyle{\cfrac { 24\sqrt { x } +7\sqrt { y } }{ \sqrt { xy } } + 31}$

By C-S inequality

$\displaystyle{\cfrac { 24 \sqrt { x } +7 \sqrt { y } }{ \sqrt { xy } }} = \displaystyle{\cfrac { 24}{ \sqrt { y }} + \cfrac{7}{ \sqrt { x }} } \leq \displaystyle{\sqrt{24^2+7^2} \sqrt{\frac{1}{y} + \frac{1}{x}} = 25 \sqrt{ \frac{x+y}{xy} } = 25}$

hence

$\displaystyle{\cfrac { 24\sqrt { x } +7\sqrt { y } +31\sqrt { xy } }{ \sqrt { xy } } } \leq 25 + 31 = 56$

Deepanshu Gupta
Mar 20, 2015

Well This is Similar Problem that I had Solved a while ago which is Posted by our friend Megh Choksi !

$Hint-1$ Given equation is equivalent to this :

$x+y=xy$

$Hint-2$ : This is represent an Right angled Triangle , because of Pythagoras theorem ! :)

Or we can use that we have to maximise

$\frac {24}{ \sqrt y} + \frac {7}{ \sqrt x} +31$

under the constraints

$\frac {1}{y} + \frac {1}{x} =1$ (which we get from first condition

let us put $\frac {1}{y} = {(sin( \theta ) )}^{2} and \frac {1}{x} = {(cos( \theta ))}^{2}$

substituting on what we have to maximise , we get

$24 cos( \theta ) + 7 sin( \theta ) + 31$ whose maximum value is 25+31 = 56

Nice problem though

Mvs Saketh - 6 years, 2 months ago

Yup! I cast it into a circle centered at origin too. Suddenly realized the importance of $24$ and $7$ ... :)

Raghav Vaidyanathan - 6 years, 2 months ago

Yes Exactly bhai ! Actually I take Inspiration from your Solution to this beautiful Problem ! :)

Megh choksi Problem-Maximum(Manupilation)

Karan Shekhawat - 6 years, 2 months ago

Yes , I remember that question !

A Former Brilliant Member - 6 years, 2 months ago

So how do you solved it ?

Karan Shekhawat - 6 years, 2 months ago

@Karan Shekhawat – Hey can anybody please tell me what are the possible values of x and y that can give 56 as the answer from the arrived equarion xy=x+y. Thanks

Rajath Naik - 6 years, 2 months ago

@Rajath Naik – Take

$\displaystyle{ x = \cfrac{625}{49}, \ \ y = \cfrac{625}{576}}$

and they realize the maximum $56$ .

You can duduce that $xy = x + y$ through straight calulation from

$\displaystyle{{ x }+{ y }={ \left( \cfrac { \sqrt { x } }{ \sqrt { y } } +\cfrac { \sqrt { y } }{ \sqrt { x } } \right) }^{ 2 }}$

Andrea Palma - 6 years, 2 months ago

@Andrea Palma – Thank u very much.I really appreciate it.I was wondering if among integers x and y can only be either 0 or 2.??

Rajath Naik - 6 years, 2 months ago

@Rajath Naik – You are welcome!

And YES! The equation $x+ y = xy$ among integers leads only to two possible consequences: $x$ and $y$ can only be both $0$ or both $2$ .

(Here's a LONG, yet rigorous, but tedious proof of this simple fact) You can think in this way.

Obviously from the equation $x = 0 \Leftrightarrow y=0$ and $x=y=0$ is indeed a solution.

So in the rest we can assume both variables are non zero.

Rearranging the equation we get $x = (x-1)y$ and $y = (y-1)x$ so we have $y$ is a divisor of $x$ and viceversa.

This means (since we are NOT considering zero values) that $x = \pm y$ .

If $x = y$ we get $2x = x^2$ that means $x(2 -x) = 0$ and finally $x = 2$ . So we have a solution $x=y=2$

If $x = - y$ we get $x^2 = 0$ that leads to nothing (assuming non zero values).

So we only got two possibilities for a solution among integers.

Namely $(x,y) = (0,0)$ and $(x,y) = (2,2)$ .

Andrea Palma - 6 years, 2 months ago

@Andrea Palma – Your explanations are good , sir :)

A Former Brilliant Member - 6 years, 2 months ago

@A Former Brilliant Member – Thanks so much! You are very kind!

Andrea Palma - 6 years, 2 months ago

@Karan Shekhawat – Well it's basically the same method used by Simran , just that I solved most parts mentally.

A Former Brilliant Member - 6 years, 2 months ago

also @Deepanshu Gupta - i think given equation is x+y = xy instead

Mvs Saketh - 6 years, 2 months ago

ohh sorry bro , actually I solved It orally , So while typing I messed up question and solution and mistype it . Thanks I have corrected It.

Deepanshu Gupta - 6 years, 2 months ago

Algebraic/Geometric Manipulations

Megh choksi

The answer is 56.

2 solutions

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