Question

If ${\log }_{10}(x^3+y^3)-{\log }_{10}(x^2+y^2-xy)\le 2$. Where $x$, $y$ are positive real numbers. Then the maximum value of ${}^{\prime }x{y}^{\prime }$ is

• A. $2500$
• B. $3000$
• C. $1200$
• D. $3500$

Answer 1

The correct option is A $2500$

${\log }_{10}\left|\frac{(x+y)(x^2+y^2-xy)}{x^2+y^2-xy}\right|\le 2{\log }_{10}|x+y|\le 2x+y\le {10}^2$

We know $AM\ge GM$

$\frac{x+y}{2}\ge (xy{)}^{1/2}\frac{(x+y{)}^2}{4}\ge xy$

Maximum value of $xy\le \frac{({10}^2{)}^2}{4}xy\le 2500$