A calculus problem by Kshetrapal Dashottar

Calculus Level 2

What is the minimum value of a a , such that for all positive real numbers x x , 4 a x 2 + 1 x ≥ 1 ? 4ax^2 + \frac{1}{x} \geq 1 ?


The answer is 0.037.

Kshetrapal Dashottar by Kshetrapal Dashottar , 20, India

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1 solution

For the minimum value of 54a .a should be minimum lets consider a function f(x)=4ax²+1/x f'(x)=8ax-1/x²≥1 so x≥1/(8a)^1/3 so f(1/(8a)^1/3) will give minimum value of a as 1/27=0.037

0 Helpful 0 Interesting 0 Brilliant 0 Confused

The question is somewhat ambiguous.

I suspect you want x x to be "any real number", instead of "Let x x be a (fixed) real number".

Calvin Lin Staff - 4 years, 6 months ago

Where does 54 come from? Why can't a be 1? If x and a were both 1 the function would equal 5 and be satisfied. If x were 1 and a were an infinitely small fraction (0.000...1) then the function would equal 1 and still be satisfied.

Joe Freeman - 4 years, 6 months ago

As reported by Jon, you have to restrict the domain to positive numbers. Do you see how this is reflected in your solution?

Calvin Lin Staff - 4 years, 6 months ago

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