CAT 2008: Quantitative Questions a Day 51-100 -The discussions-Brought to you by SMOT

arbit_rageur

2>=p-q

So 2+q>=p. => 2/(p+q)+q/2>=2/(2+2q)+q/2>=sqrt(2)-1.

So (1).

Aarav

Can all those who are claiming the answer as 1/√2 tell the corresponding values of p and q?

ljv

Here is my take,

2/(p+q) +q/2 is going to be minimum when, 2/(p+q) takes min value and q/2 takes min value.

This implies q be as low as possible and (p+q) as high as possible subject to the condition p>0 and q>0. & 2>= p-q.

Therefore we let q to tend to zero, ie take values very near to zero say 0.000000001 this implies that p will take values very close to 2 but less than 2 like 1.999999999.

Then 2/(p+q) will give the value = 1
and q/2 will give values tending to zero like 0.0000000005
giving the sum as 1.0000000005
thus the minima of the sum tending to 1.

Therefore my answer is option(3) ie 1

Aarav

Quote:

Originally Posted by arbit_rageur

2>=p-q

So 2+q>=p. => 2/(p+q)+q/2>=2/(2+2q)+q/2>=sqrt(2)-1.

So (1).

This is OK - you have proved here that our expression is >= sqrt(2) - 1, but is this the minimum value only? Just for the thought!

arbit_rageur

Yeah, this is the minimum value only...p=sqrt(2)+1 and q=sqrt(2)-1 will give us the minimum.

Quote:

Originally Posted by Aarav

This is OK - you have proved here that our expression is >= sqrt(2) - 1, but is this the minimum value only? Just for the thought!

Aarav

@arbit_rageur
A better way to do what you have done would be

p+k = 2+q where k >= 0
Thus, 2/(p+q) + q/2 = 2/(2+2q-k) + (2q+2-k)/4 + (k-2)/4.
By AM-GM on positive integers, q can be generated based on k
and our expression is >= sqrt(2) + (k-2)/4 which assumes min at k = 0.

I would have done this way. Your thoughts on the same?

mehrozkhn

thanx for pointing out the mistake buddy . My post stands edited, plz take a look

Quote:

Originally Posted by shivam_01

hey u have taken that one wrong
p-q <= 2
=> (p+q)^2 - 2pq <= 4
this should be written as
(p+q)^2-4pq <= 4

pnvishal

QQAD 51,

<FONT face=Georgia>

arbit_rageur

Your method is ofcourse perfect and intuitively more understandable.

Only thing is IMO,we would need to add one more step after finding k=0, i.e find q and make sure it is a positive value for which the minimum is attained...ofcourse in this case, it is very obvious...but in more complicated cases, when having two variables and minimizing two parts of the expression separately ,it may become an issue. ....Your thoughts?

Quote:

Originally Posted by Aarav

@arbit_rageur
A better way to do what you have done would be

p+k = 2+q where k >= 0
Thus, 2/(p+q) + q/2 = 2/(2+2q-k) + (2q+2-k)/4 + (k-2)/4.
By AM-GM on positive integers, q can be generated based on k
and our expression is >= sqrt(2) + (k-2)/4 which assumes min at k = 0.

I would have done this way. Your thoughts on the same?

ravishah17

Given that

p>0
q>0
p-q<=2

Min value of f=2/(p+q)+q/2

Have approached using LPP,

Plotted a graph of three inequalities,
Found cut off points at p=0,q=0 & p=2,q=0

For (0,0); Value of f is infinity
For (2,0); Value of f is 1

My answer (3) 1

Have just now seen other answers, don't know which approach is correct
I think using LPP is wrong as values considered are only integers...