Question

If for two positive integers $x,y$ with $xy=16$ then find the minimum value of $x+y$.

Answer 1

Given for two positive integers $x,y$ with $xy=16$......(1).

Let $s=x+y$

or, $s=x+\frac{16}{x}$ [ From (1)]

Now $\frac{ds}{dx}=1-\frac{16}{x^2}$

And $\frac{d^{2}s}{d{x}^2}=\frac{32}{x^3}$.

For maximum or minimum value of $s$ we must have $\frac{ds}{dx}=0$.

Then $1-\frac{16}{x^2}=0$

or, $x^2=16$

or, $x=\pm 4$.

But at $x=4$ we have $\frac{d^{2}s}{d{x}^2}>0$.

So for $x=4$ we get minimum value of $s$.

If $x=4$ then $y=4$ from (1).

So minimum value of $s$ is $4+4=8$.