Question

Mark the correct alternative of the following. Let $x,y$ be two variables and \[x>0,xy=1\], then the minimum value of $x+y$ is?
A. 1
B. 2
C. $2\dfrac{1}{2}$
D. $3\dfrac{1}{3}$

Answer 1

Hint: We will differentiate the given function $x+y$ for which we have to find the minimum value. First, we have to find critical points by differentiating a function once and equating it to zero. We will further differentiate the function to find at which critical point the function is minimum and then we will substitute the critical point in function to find the minimum value.

Complete step-by-step answer:

Given, $xy=1\Rightarrow y=\dfrac{1}{x}\cdot \cdot \cdot \cdot \left( 1 \right)$.
We have to find the minimum value of $x+y$, now we will substitute equation 1 in $x+y$ making the function as a uniform variable. So, it becomes $x+\dfrac{1}{x}$ and let it be $p$.
To find the critical points of a function, we have to differentiate the function once and then equate it to zero. Now, we will find critical points by differentiating $p$ with respect to $x$ and equating to zero.
$\dfrac{dp}{dx}=\dfrac{d\left( x+\dfrac{1}{x} \right)}{dx}=1-\dfrac{1}{{{x}^{2}}}$ as $\dfrac{d\left( \dfrac{1}{x} \right)}{dx}=\dfrac{-1}{{{x}^{2}}}$ and $\dfrac{dx}{dx}=1$
$\dfrac{dp}{dx}=1-\dfrac{1}{{{x}^{2}}}\cdot \cdot \cdot \cdot (2)$
Equating to zero, we get
$\begin{align}
  & \Rightarrow 1-\dfrac{1}{{{x}^{2}}}=0 \\
 & \Rightarrow \dfrac{1}{{{x}^{2}}}=1 \\
 & \Rightarrow {{x}^{2}}=1 \\
 & \Rightarrow x=\pm 1 \\
\end{align}$
So, $x=1$ and $x=-1$ are critical points.
Now to get the point at which the function will have a minimum or maximum value, we have to differentiate it twice and substitute the critical points in it. If the value hence obtained is less than 0, then we know that the function will have a maximum value and if the value obtained is greater than 0, then the function will have a minimum value.
Now, we will double differentiate the function and substitute critical points to find at which point the function is minimum.
$\dfrac{{{d}^{2}}p}{d{{x}^{2}}}=\dfrac{d\left( \dfrac{dp}{dx} \right)}{dx}$
From equation 2, we have $\dfrac{dp}{dx}=1-\dfrac{1}{{{x}^{2}}}$.
$\Rightarrow \dfrac{{{d}^{2}}p}{d{{x}^{2}}}=\dfrac{d\left( 1-\dfrac{1}{{{x}^{2}}} \right)}{dx}\Rightarrow \dfrac{2}{{{x}^{3}}}\cdot \cdot \cdot \cdot (3)$ as $\dfrac{d\left( \dfrac{1}{{{x}^{2}}} \right)}{dx}=\dfrac{-2}{{{x}^{3}}}$
So, $\dfrac{{{d}^{2}}p}{d{{x}^{2}}}=\dfrac{2}{{{x}^{3}}}$. Now we will substitute critical points $x=1$ and $x=-1$ in equation 3. If $\dfrac{{{d}^{2}}p}{d{{x}^{2}}}>0$, then the function will be minimum at this critical point.
Let us take $x=-1$, in this case $\dfrac{{{d}^{2}}p}{d{{x}^{2}}}=\dfrac{2}{{{x}^{3}}}=\dfrac{2}{{{(-1)}^{3}}}=-2$, here it is less than 0. So, we have to check the other point as it is not a valid critical point.
Now, let us take $x=1$, in this case $\dfrac{{{d}^{2}}p}{d{{x}^{2}}}=\dfrac{2}{{{x}^{3}}}=\dfrac{2}{{{(1)}^{3}}}=2$, here it is greater than 0. So, at $x=1$ the function is minimum.
So, we will substitute $x=1$ in the function $p$.
$p=x+\dfrac{1}{x}=1+\dfrac{1}{1}=2$
So, the minimum value of $x+y$ is 2.
Hence, the correct option is B.

Note: This problem can be solved in an alternative way. It is given that $xy=1$ and we have to find the minimum value of $x+y$. The function will be minimum if both $x$ and $y$ are equal in equation 3, i.e. $x=y$. So, function becomes ${{x}^{2}}=1\Rightarrow x=\pm 1$. But we need to have $x>0$ according to the condition in the question, so, $x=1$. Now we will find $y$ by substituting the value of $x$ as,
$\begin{align}
  & \Rightarrow (1)y=1 \\
 & \Rightarrow y=1 \\
\end{align}$
$\Rightarrow 1+1=2$
Hence the minimum value of $x+y$ is 2 and the correct option is B.