Question

If ${\log }_{10}(x^3+y^3)-{\log }_{10}(x^2+y^2-xy)\le 2$, then the maximum value of $xy$, for all $x\ge 0,y\ge 0$, is

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

Answer 1

The correct option is A $2500$

$\Rightarrow {\log }_{10}\left(\frac{x^3+y^3}{x^2+y^2-xy}\right)\le 2$
$\Rightarrow {\log }_{10}\left[\frac{(x+y)(x^2+y^2)-xy}{x^2+y^2-xy}\right]\le 2$
${\log }_{10}(x+y)\le 2\Rightarrow (x+y)\le {10}^2=100$
since A.M $\ge$ G.M
$\Rightarrow \frac{x+y}{2}\ge \sqrt{xy}\Rightarrow \sqrt{xy}\le \frac{x+y}{2}\le 50$
$\Rightarrow xy\le 2500\Rightarrow Max.\left(xy\right)=2500$