Question

Let x, y be two variables and x>0, xy=1, then minimum value of x+y is
(a) 1

(b) 2

(c) $2\frac{1}{2}$

(d) $3\frac{1}{3}$

Answer 1

(b) 2

$$\begin{gathered} \mathrm{Given}:xy=1 \\ \Rightarrow y=\frac{1}{x} \\ f\left(x\right)=x+\frac{1}{x} \\ \Rightarrow f\text{'}\left(x\right)=1-\frac{1}{x^2} \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow 1-\frac{1}{x^2}=0 \\ \Rightarrow x^2-1=0 \\ \Rightarrow x^2=1 \\ \Rightarrow x=\pm 1 \\ \Rightarrow x=1\left(\mathrm{Given}:\text{x}>1\right) \\ \Rightarrow y=1 \\ \mathrm{Now}, \\ f\text{'}\text{'}\left(x\right)=\frac{2}{x^3} \\ \Rightarrow f\text{'}\text{'}\left(1\right)=2>0 \\ \mathrm{So},x=1\mathrm{is}\mathrm{a}\mathrm{local}\text{minima}. \\ \therefore \mathrm{Minimum}\mathrm{value}\mathrm{of}f\left(x\right)=f\left(1\right)=1+1=2 \end{gathered}$$