Question

Answer the following questions in one word or one sentence or as per exact requirement of the questions.
Find the least value of $f\left( x \right) = ax + \dfrac{b}{x}$, where a>0, b>0 and x>0.

Answer 1

Hint- In this question use the concept of first derivative, simply take the first derivative of the given function and equate it to 0, this will give the value of x in terms of a and b. Then check for double derivative rules to make sure that the value of x obtained, corresponds to the least value of f(x). This will help approaching the files.

Complete step-by-step solution -

Given expression:
$f\left( x \right) = ax + \dfrac{b}{x}$
Now to find the least value of the given expression we have to differentiate it w.r.t. x and equate to zero, so we have,
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}\left( {ax + \dfrac{b}{x}} \right) = 0$
Now differentiate according to the property $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ so we have,
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = a - \dfrac{b}{{{x^2}}} = 0$
$ \Rightarrow a = \dfrac{b}{{{x^2}}}$
$ \Rightarrow {x^2} = \dfrac{b}{a}$
$ \Rightarrow x = \sqrt {\dfrac{b}{a}} $
Now double differentiate the given equation we have,
$ \Rightarrow \dfrac{{{d^2}}}{{d{x^2}}}f\left( x \right) = \dfrac{{{d^2}}}{{d{x^2}}}\left( {ax + \dfrac{b}{x}} \right) = \dfrac{d}{{dx}}\left( {a - \dfrac{b}{{{x^2}}}} \right) = 0 - \dfrac{{ - 2b}}{{{x^3}}} = \dfrac{{2b}}{{{x^3}}}$
Now substitute the value of x in above equation we have,
$ \Rightarrow \dfrac{{{d^2}}}{{d{x^2}}}f\left( x \right) = \dfrac{{2b}}{{{x^3}}} = \dfrac{{2b}}{{{{\left( {\sqrt {\dfrac{b}{a}} } \right)}^3}}}$
So as we see that the double differentiate is positive at $x = \sqrt {\dfrac{b}{a}} $ so the function f(x) has minimum value at this value of x, if we got double differentiate negative at $x = \sqrt {\dfrac{b}{a}} $ than the function has maximum value at this value of x.
So the minimum value of x has
$ \Rightarrow f{\left( x \right)_{\min }} = ax + \dfrac{b}{x} = a\sqrt {\dfrac{b}{a}} + \dfrac{b}{{\sqrt {\dfrac{b}{a}} }}$
Now simplify this we have,
$ \Rightarrow f{\left( x \right)_{\min }} = a\sqrt {\dfrac{b}{a}} + \dfrac{b}{{\sqrt {\dfrac{b}{a}} }} = a\sqrt {\dfrac{b}{a}} + b\sqrt {\dfrac{a}{b}} $
$ \Rightarrow f{\left( x \right)_{\min }} = \sqrt {\dfrac{{{a^2}b}}{a}} + \sqrt {\dfrac{{a{b^2}}}{b}} = \sqrt {ab} + \sqrt {ab} = 2\sqrt {ab} $
So this is the required least value of the given expression.

Note – The double derivative rule helps finding whether the value corresponds to x, enables least value or function or the max value of the function. If the value of double derivative thus obtained is negative then the value corresponds to the max value, however if it is positive then the value of x obtained corresponds to the minima of the function. Thus the conditions given on a and b is very important as in this question a>0, b>0 and x>0.