Question

If $\begin{array}{rl}\frac{1}{x}+\frac{1}{y}& =1\end{array}$, the function $a^3{x}^2+b^3{y}^2\begin{array}{}\end{array}$ is minimum, then $x$ and $y$ are

• A. $a+b$, $a-b$
• B. $a$,$b$
• C. $\frac{a}{a+b}$,$\frac{b}{a+b}$
• D. $\frac{a+b}{a}$, $\frac{a+b}{b}$

Answer 1

The correct option is D

$\frac{a+b}{a}$, $\frac{a+b}{b}$

Step 1. Given data:

$\begin{array}{rl}\Rightarrow \frac{1}{x}+\frac{1}{y}& =1\\ \frac{1}{y}& =1-\frac{1}{x}\\ \frac{1}{y}& =\frac{x-1}{x}\end{array}$

Step 2. Calculate the value of $y$

$\begin{array}{rl}\Rightarrow \frac{1}{x}+\frac{1}{y}& =1\\ \frac{1}{y}& =1-\frac{1}{x}\\ \frac{1}{y}& =\frac{x-1}{x}\end{array}$

Taking reciprocal

$y=\frac{x}{x-1}$

$\begin{array}{rl}Letf\left(x\right)& =a^3{x}^2+b^3{y}^2\\ & =a^3{x}^2+b^3(\frac{x}{x-1}{)}^2\end{array}$

Step 3. Differentiate $f(x)$ with respect to $x$

$\begin{array}{rl}\frac{df}{dx}& =2a^{3}x+2b^3\frac{x}{x-1}\left(\frac{x-1-x}{(x-1{)}^2}\right)\\ & =2a^{3}x-\frac{2b^{3}x}{(x-1{)}^3}\end{array}$

Step 4. For minima of $f(x)$, put $\frac{df}{dx}=0$

$\begin{array}{ll}0& =2a^{3}x-\frac{2b^{3}x}{(x-1{)}^3}\\ \Rightarrow 0& =2x(a^3-\frac{b^3}{(x-1{)}^3})\\ \Rightarrow 0& =a^3-\frac{b^3}{(x-1{)}^3}\\ \Rightarrow a^3& =\frac{b^3}{(x-1{)}^3}\\ \Rightarrow (x-1{)}^3& =\frac{b^3}{a^3}\\ \Rightarrow (x-1)& =\frac{b}{a}\\ \Rightarrow x& =\frac{b}{a}+1\\ \Rightarrow x& =\frac{b+a}{a}\end{array}$

Step 5. Find the value of $y$

Put $x=\frac{b+a}{a}$ in $y=\frac{x}{x-1}$

$\begin{array}{rl}y& =\frac{\frac{b+a}{a}}{\frac{b}{a}}\\ & =\frac{b+a}{b}\end{array}$

Hence, Option(D) is the correct option.