Question

if x varies inversely with y then
prove that the minimum value of (x+y) is constant if x=y
​

Answer 1

Given data:

$\mathsf{x}$ varies inversely with $\mathsf{y}$

To prove:

$\mathsf{(x+y)_{min}=constant}$ if $\mathsf{x=y}$

Step-by-step explanation:

Given, $\mathsf{x\propto \frac{1}{y}}$

$\mathsf{\Rightarrow x=\frac{k}{y}}$ where $\mathsf{k}$ is constant

$\mathsf{\Rightarrow xy=k}$

$\mathsf{\Rightarrow x\times x=k}$ when $\mathsf{x=y}$

$\mathsf{\Rightarrow x^{2}=k}$

$\mathsf{\Rightarrow x=\sqrt{k}}$

Then $\mathsf{y=\sqrt{k}}$

Now, $\mathsf{x+y=\sqrt{k}+\sqrt{k}}$

$\mathsf{\Rightarrow x+y=2\sqrt{k}}$, a constant

Answer:

The minimum value of $\mathsf{(x+y)}$ is constant.

Hence, proved.

Another method.

We can prove the problem using

$\mathsf{\quad A.M.\geq G.M.}$

$\mathsf{\Rightarrow \frac{x+y}{2}\geq \sqrt{xy}}$

$\mathsf{\Rightarrow \frac{x+y}{2}\geq \sqrt{k}}$ since $\mathsf{xy=k}$

$\mathsf{\Rightarrow x+y\geq 2\sqrt{k}}$

Here $\mathsf{k}$ being a constant, $\mathsf{2\sqrt{k}}$ is also a constant.

$\mathsf{\therefore}$ the minimum value of $\mathsf{(x+y)}$ is constant.