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

Answer 1

The correct option is B

$2500$

Explanation for the correct option:

Simplifying ${\mathbf{log}}_{\mathbf{10}}\mathbf{}\mathbf{(}{\mathbf{x}}^{\mathbf{3}}\mathbf{+}{\mathbf{y}}^{\mathbf{3}}\mathbf{)}\mathbf{–}{\mathbf{log}}_{\mathbf{10}}\mathbf{}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{–}\mathbf{xy}\mathbf{)}\mathbf{\le }\mathbf{}\mathbf{2}$:

$\begin{array}{rcl}{\log }_{10}(x^3+y^3)–{\log }_{10}(x^2+y^2–xy)& =& {\log }_{10}\left(\frac{(x^3+y^3)}{(x^2+y^2-xy)}\right)\left[\mathrm{by}\log a-\log b=\log \left(\frac{a}{b}\right)\right]\\ & =& {\log }_{10}\left(\frac{\left(x+y\right)\left(x^2+y^2-xy\right)}{x^2+y^2-xy}\right)\left[\mathrm{by}{a}^3+b^3=(a+b)(a^2-ab+b^2)\right]\\ & =& {\log }_{10}\left(x+y\right)\end{array}$

$$\begin{gathered} \Rightarrow {\log }_{10}\left(x+y\right)\le 2 \\ \Rightarrow x+y\le {10}^2 \\ \Rightarrow x+y\le 100 \end{gathered}$$

Finding the value of $\mathit{x}\mathit{y}$:

It is given that $x\ge 0,y\ge 0$, so the arithmetic mean of $x\mathrm{and}y\ge$ their geometric mean.

$$\begin{gathered} \Rightarrow \frac{x+y}{2}\ge \sqrt{xy}byA.M=\frac{\underset{}{\sum n}}{n}andG.M=\sqrt[n]{\mathrm{Product}\mathrm{of}n\mathrm{observations}} \\ \Rightarrow 50\ge \sqrt{xy} \end{gathered}$$

By squaring both sides, we get

$xy\le 2500$

Hence, the correct answer is option B.