Question

For any $x,y\in R$, xy > 0 then the minimum value of $\dfrac{2x}{{{y}^{3}}}+\dfrac{{{x}^{3}}y}{3}+\dfrac{4{{y}^{2}}}{9{{x}^{4}}}$ is equal to
(a) ${{2}^{\dfrac{1}{3}}}$
(b) 2
(c) ${{3}^{\dfrac{1}{3}}}$
(d) 3

Answer 1

First, before proceeding for this, we must know the concept of Arithmetic Mean(AM) and Geometric Mean(GM) which gives the following condition as $AM\ge GM$. Then, Arithmetic Mean(AM) is the mean of the terms which means the summation of the terms divided by the number of the terms and Geometric Mean(GM) is the multiplication of the given terms with raise to the power of the reciprocal of the number of terms. Then, by using the condition $AM\ge GM$ and substituting the value of AM and GM, we get the minimum value of the function.

Complete step-by-step solution:
In this question, we are supposed to find the minimum value of the function $\dfrac{2x}{{{y}^{3}}}+\dfrac{{{x}^{3}}y}{3}+\dfrac{4{{y}^{2}}}{9{{x}^{4}}}$ for which the condition is given as $x,y\in R$, xy>0.
So, before proceeding for this, we must know the concept of Arithmetic Mean(AM) and Geometric Mean(GM) which gives the following condition as:
$AM\ge GM$
Now, before applying the above condition, we must know about the AM and GM.
So, Arithmetic Mean(AM) is the mean of the terms which means the summation of the terms divided by the number of the terms.
Then, we get AM for the above equation as:
$AM=\dfrac{\dfrac{2x}{{{y}^{3}}}+\dfrac{{{x}^{3}}y}{3}+\dfrac{4{{y}^{2}}}{9{{x}^{4}}}}{3}$
Now, Geometric Mean(GM) is the multiplication of the given terms with raise to the power of the reciprocal of the number of terms.
Then, we get GM for the above equation as:
$GM={{\left( \dfrac{2x}{{{y}^{3}}}\times \dfrac{{{x}^{3}}y}{3}\times \dfrac{4{{y}^{2}}}{9{{x}^{4}}} \right)}^{\dfrac{1}{3}}}$
Now, by using the condition $AM\ge GM$ and substituting the value of AM and GM, we get:
$\dfrac{\dfrac{2x}{{{y}^{3}}}+\dfrac{{{x}^{3}}y}{3}+\dfrac{4{{y}^{2}}}{9{{x}^{4}}}}{3}\ge {{\left( \dfrac{2x}{{{y}^{3}}}\times \dfrac{{{x}^{3}}y}{3}\times \dfrac{4{{y}^{2}}}{9{{x}^{4}}} \right)}^{\dfrac{1}{3}}}$
So, the above equation can give us the minimum value of the function which is on the left side of the expression.
Then, multiply the both sides by 3 and then by solving, we get:
$\dfrac{2x}{{{y}^{3}}}+\dfrac{{{x}^{3}}y}{3}+\dfrac{4{{y}^{2}}}{9{{x}^{4}}}\ge 3\times {{\left( \dfrac{2x}{{{y}^{3}}}\times \dfrac{{{x}^{3}}y}{3}\times \dfrac{4{{y}^{2}}}{9{{x}^{4}}} \right)}^{\dfrac{1}{3}}}$
Now, cancel the terms in the right hand side to get the minimum value of the function as:
$\begin{aligned} & \dfrac{2x}{{{y}^{3}}}+\dfrac{{{x}^{3}}y}{3}+\dfrac{4{{y}^{2}}}{9{{x}^{4}}}\ge 3\times {{\left( \dfrac{8{{x}^{4}}{{y}^{3}}}{27{{x}^{4}}{{y}^{3}}} \right)}^{\dfrac{1}{3}}} \\\ & \Rightarrow \dfrac{2x}{{{y}^{3}}}+\dfrac{{{x}^{3}}y}{3}+\dfrac{4{{y}^{2}}}{9{{x}^{4}}}\ge 3\times {{\left( \dfrac{8}{27} \right)}^{\dfrac{1}{3}}} \\\ & \Rightarrow \dfrac{2x}{{{y}^{3}}}+\dfrac{{{x}^{3}}y}{3}+\dfrac{4{{y}^{2}}}{9{{x}^{4}}}\ge 3\times \dfrac{2}{3} \\\ & \Rightarrow \dfrac{2x}{{{y}^{3}}}+\dfrac{{{x}^{3}}y}{3}+\dfrac{4{{y}^{2}}}{9{{x}^{4}}}\ge 2 \\\ \end{aligned}$
So, from the above expression, it is clear that the minimum value of the function $\dfrac{2x}{{{y}^{3}}}+\dfrac{{{x}^{3}}y}{3}+\dfrac{4{{y}^{2}}}{9{{x}^{4}}}$ is 2.
Hence, option (b) is correct.

Note: Now, to solve these types of questions we need to know some of the basic rules for the exponents functions so that they can be solved easily. So, the rules used in above question are as:
$\begin{aligned} & {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} \\\ & \dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}} \\\ & {{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}} \\\ \end{aligned}$
So, these rules are kept in mind while solving these types of problems.
Sometimes students think to solve this problem by using the second derivative method to find the minimum value but that would be a long approach as the given function has two variables so this would be complex to find the derivative.