Question

If ${\log }_{10}\left(x^3+y^3\right)-{\log }_{10}\left(x^2+y^2-xy\right)\le 2$, 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 answer:

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

$$\begin{gathered} \Rightarrow {\log }_{10}\frac{\left(x^3+y^3\right)}{\left(x^2+y^2-xy\right)}\le 2\mathbf{}\left[\because loga-logb=log\frac{a}{b}\right] \\ \Rightarrow {\log }_{10}\left(\frac{(x+y)(x^2+y^2-xy)}{\left(x^2+y^2-xy\right)}\right)\le 2\left[\because (x^3+y^3)=(x+y)(x^2+y^2-xy)\right] \\ \Rightarrow {\log }_{10}\left(x+y\right)\le 2 \\ \Rightarrow (x+y)\le {10}^2\left[\because lo{g}_a\left(b\right)=c\Rightarrow b=a^c\right] \\ \Rightarrow x+y\le 100 \end{gathered}$$

We know that AM$\ge$GM where AM is Arithmetic mean and GM is Geometric mean

$$\begin{gathered} \Rightarrow \sqrt{xy}\le \frac{x+y}{2} \\ \Rightarrow \sqrt{xy}\le \frac{100}{2}\left[\because (x+y)\le 100\right] \\ \Rightarrow \sqrt{xy}\le 50 \\ \Rightarrow xy\le 2500 \end{gathered}$$

Maximum value of $xy=2500$

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