Because x,y integer so x+y integer so x+y \(\le39\)

We have:\(\left(x-y\right)^2\ge0\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow x^2+2xy+y^2\ge4xy\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Rightarrow xy\le\dfrac{\left(x+y\right)^2}{4}\le\dfrac{38^2}{4}=361\)(Because \(39^2⋮̸4\))

So maximum of xy is 361 when x=y=19

error:wrong question

Error, sr

We have: x + y < 40 

=> \(x+y\le39\)

=> \(\left(x+y\right)^2\le1521\)

=> \(x^2+2xy+y^2\le1521\)

=> \(xy\le\dfrac{1521-x^2-y^2}{2}\)