It's just a minimum

Calculus Level 3

If a , b a,b and c c are constants satisfying the inequality a x 2 + b x ≥ c ax^2 + \dfrac bx \geq c , where x > 0 x> 0 , find the minimum vlaue of a b 2 c 3 \dfrac{ab^2}{c^3} .

If the minimum value can be expressed as p q \dfrac pq , where p p and q q are coprime positive integers, submit your answer as p + q p+q .

This one of the old IIT question.


The answer is 31.

Rishi Sharma by Rishi Sharma , 32, 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.

1 solution

Rishi Sharma
May 26, 2016

First notice that

a x 2 + b x = a x 2 + b 2 x + b 2 x a{x}^{2}+\frac{b}{x}\ = a{x}^{2}+ \frac{b}{2x}+\frac{b}{2x}

Now apply AM-GM a x 2 + b 2 x + b 2 x 3 ≥ ( a b 2 4 ) 1 3 \frac{a{x}^{2}+ \frac{b}{2x}+\frac{b}{2x}}{3}\ge { \left( \frac { a{ b }^{ 2 } }{ 4 } \right) }^{ \frac { 1 }{ 3 }} If we set 3 ( a b 2 4 ) 1 3 ≥ c 3{ \left( \frac { a{ b }^{ 2 } }{ 4 } \right) }^{ \frac { 1 }{ 3 }}\ge c Than our original equation will always remain true. Hence 27 ( a b 2 4 ) ≥ c 3 27{ \left( \frac { a{ b }^{ 2 } }{ 4 } \right) }\ge {c}^{3} or a b 2 c 3 ≥ 4 27 \frac{a{b}^{2}}{{c}^{3}}\ge \frac{4}{27} Hence our answer is 4+27= 31

There is also a calculus approach.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

If we set 3 ( a b 2 4 ) 1 3 ≥ c 3{ \left( \frac { a{ b }^{ 2 } }{ 4 } \right) }^{ \frac { 1 }{ 3 }}\ge c . Than our original equation will always remain true.

More accurately, it should be phrased as "Thus, the original equation is true if and only if 3 ( a b 2 4 ) 1 3 ≥ c 3{ \left( \frac { a{ b }^{ 2 } }{ 4 } \right) }^{ \frac { 1 }{ 3 }}\ge c ".

Otherwise, as written, you have found a sufficient condition but haven't explained that it is necessary. Thus, the minimum bound that you have applies to this restricted set (with a sufficient condition), and need not apply to the possibly larger set of solutions.

As an explicit example, "If we set ( a b 2 4 ) 1 3 ≥ c { \left( \frac { a{ b }^{ 2 } }{ 4 } \right) }^{ \frac { 1 }{ 3 }}\ge c . Than our original equation will always remain true." This would then give us a b 2 c 3 ≥ 4 \frac{ ab^2 } { c^3 } \geq 4 which clearly isn't the minimium of all possible values.

Did the same. the same kind of question was there in JEE 2016

Prakhar Bindal - 5 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...