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

$2500$

Explanation for the correct option

The given expression is

${\log }_{10}(x^3+y^3)-{\log }_{10}(x^2+y^2-xy)\le 2$

$\Rightarrow$ ${\log }_{10}\left(\frac{x^3+y^3}{x^2+y^2-xy}\right)\le 2$ $\left[\because {\log }_{a}m-{\log }_{a}n={\log }_a\left(\frac{m}{n}\right)\right]$

Factorizing $x^3+y^3$ we get

$\Rightarrow$ ${\log }_{10}\left(\frac{\left(x^2+y^2-xy\right)\left(x+y\right)}{x^2+y^2-xy}\right)\le 2$ $\left[\because x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\right]$

$\Rightarrow$ ${\log }_{10}\left(x+y\right)\le 2$

$\Rightarrow$ $x+y\le {10}^2$ $\left[\because {\log }_{a}b=x\Rightarrow b=a^x\right]$

$\Rightarrow$ $x+y\le 100$ $...\left(i\right)$

From the A.M-G.M inequality we know that

A.M$\ge$G.M

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

$\Rightarrow$$\frac{100}{2}\ge \sqrt{xy}$ [From $\left(i\right)$]

$\Rightarrow$ $xy\le {\left(50\right)}^2$

$\Rightarrow$ $xy\le 2500$

Thus, the maximum value of$xy$ is $2500$.

Hence, option (A) i.e. $2500$ is the correct answer.