There's only one answer?

Algebra Level 3

Given that p , q p,q and r r are positive real numbers satisfying 27 p q r ≥ ( p + q + r ) 3 27pqr \geq (p+q+r)^3 and 3 p + 4 q + 5 r = 12 3p+4q + 5r=12 , find the value of 8 p + 4 q − 7 r 8p + 4q - 7r .

If you're looking to skyrocket your preparation for JEE-2015, then go for solving this set of questions .


The answer is 5.

View wiki
Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Sandeep Bhardwaj
Feb 19, 2015

Given : 27 p q r ≥ ( p + q + r ) 3 27pqr \geq (p+q+r)^3 ------------------------------(i)

But using A.M.-G.M. inequality,

⟹ p + q + r 3 ≥ ( p q r ) 1 3 \implies \dfrac{p+q+r}{3} \geq (pqr)^{\frac{1}{3}}

⟹ ( p + q + r ) 3 ≥ 27 p q r \implies (p+q+r)^3 \geq 27pqr -------------------------------(ii)

From (i) and (ii), we can say that A . M . = G . M . A.M.=G.M. which will hold when p = q = r p=q=r .

Also, given that 3 p + 4 q + 5 r = 12 3p+4q+5r=12

⟹ p = q = r = 1 \implies p=q=r=1

Hence , 8 p + 4 q − 7 r = 5 . \boxed{8p+4q-7r=5}.

enjoy!

24 Helpful 0 Interesting 0 Brilliant 0 Confused

This question is very special for me , Since this question is asked in 11th class by my maths tution teacher , and I was the only student who got correct answer to this question , and no other would solved this .. Including JEE-2014 , AIR-187 and AIR -324 , Since these were my freinds at that time ... XD

Nishu sharma - 6 years, 1 month ago

Exactly what I did :)

Mayank Singh - 6 years, 3 months ago

Apply A.M - G.M . You will find that only equality holds. P = Q = R.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...