Application of Derivatives 4 Question 58
61. If $x$ and $y$ be two real variables such that $x>0$ and $x y=1$. Then, find the minimum value of $x+y$.
(1981, 2M)
Integer Answer Type Questions
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Answer:
Correct Answer: 61. (7)
Solution:
- Let $f(x)=x+y$, where $x y=1$
$$ \begin{array}{ll} \Rightarrow & f(x)=x+\frac{1}{x} \\ \Rightarrow & f^{\prime}(x)=1-\frac{1}{x^{2}}=\frac{x^{2}-1}{x^{2}} \\ \text { Also, } & f^{\prime \prime}(x)=2 / x^{3} \end{array} $$
On putting $f^{\prime}(x)=0$, we get
$$ x= \pm 1, \text { but } x>0 \text { [neglecting } x=-1] $$
$$ (x)>0 \text {, for } x=1 $$
Hence, $f(x)$ attains minimum at $x=1, y=1$
$\Rightarrow(x+y)$ has minimum value 2 .
