Question

Let x and y be two real numbers such that $x > 0$ and $xy = 1$. The minimum value of $x + y$ is
A) $1$
B)$\dfrac{1}{2}$
C) $2$
D)$\dfrac{1}{4}$

Answer 1

Hint: We first convert the given function in one variable (say x) then equate its first derivative to get a solution if possible and substituting the solution in the second derivative to discuss whether it is a point of minima or maxima. Finally substituting value in given equation to get corresponding maximum or minimum value of the function.

Formulas used: $\dfrac{{d{{\left( x \right)}^n}}}{{dx}} = n{\left( x \right)^{n - 1}}$ , \[\dfrac{{d\left( 1 \right)}}{{dx}} = 0,\,\,\,\dfrac{{d\left( x \right)}}{{dx}} = 1\]

Complete step-by-step solution:
 We have $xy = 1$
$ \Rightarrow y = \dfrac{1}{x}$
 Let$A = x + y$ , substituting value of $y = \dfrac{1}{x}$ we have,
$A = x + \dfrac{1}{x}$
Now differentiating above equation w.r.t. to $'x'$ we have,
$\dfrac{{dA}}{{dx}} = 1 + \left( {\dfrac{{ - 1}}{{{x^2}}}} \right)$ $\because $$\dfrac{{d\left( x \right)}}{{dx}} = 1,\,\,\dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right) = \left( {\dfrac{{ - 1}}{{{x^2}}}} \right)$

Now, equating $\dfrac{{dA}}{{dx}} = 0$ to get solution
$1 + \left( {\dfrac{{ - 1}}{{{x^2}}}} \right) = 0$
$ \Rightarrow 1 = \dfrac{1}{{{x^2}}}$
$ \Rightarrow {x^2} = 1$
$ \Rightarrow x = \pm 1$
But, it is given$x > 0$. Therefore $x = - 1$ is rejected.

To discuss whether $x = 1$is a point of maxima or minima. We calculate the second order derivative of A.
$\dfrac{{{d^2}A}}{{d{x^2}}} = 0 + \left( {\dfrac{2}{{{x^3}}}} \right)$ \[\left\{ {\dfrac{{d\left( 1 \right)}}{{dx}} = 0,\,\,\,\dfrac{d}{{dx}}\left( {\dfrac{{ - 1}}{{{x^2}}}} \right) = \dfrac{2}{{{x^3}}}} \right\}\]

Now, substituting the value of x =1 in the second derivative of A. We have
$\dfrac{{{d^2}A}}{{d{x^2}}} = \dfrac{2}{{{{\left( 1 \right)}^3}}}$
$\dfrac{{{d^2}A}}{{d{x^2}}} = 2$
Here, clearly $\dfrac{{{d^2}A}}{{d{x^2}}}$ is greater than zero or we can say positive for value $x = 1.$
Therefore, we can say that $x = 1$ is a point of minima.

Now, we used the value of $x = 1$ in step one equation to get the value of y.
i.e. $y = \dfrac{1}{x}$
or
$y = \dfrac{1}{1}$ $\left\{ {\because x = 1} \right\}$
Therefore $y = 1$ when$x = 1$.

Now, to find the minimum value of the given function $x + y$ or $A = x + y$ we substitute values of $x = 1$and $y = 1$.
$\therefore A = x + y$
$ \Rightarrow A = 1 + 1$
$ \Rightarrow A = 2$

Hence, from above we can say that the minimum value of $x + y$ is $2$ when $xy > 1$ with condition$x > 0$.

Notes: Maximum and minimum value of the function can be calculated by using derivative method. If the second derivative of the function is negative at particular point then point will be a point of maxima and if second order derivative is positive at particular point then point will be a point of minima.