Tag Archives: Inequality
$3x+4y=10$$x>0,y>0$$(x^2y^3)_{max}=?$
For positive numbers, A.M. $\geq $ G.M.
So, $\frac{{\frac{{3x}}{2} + \frac{{3x}}{2} + \frac{{4y}}{3} + \frac{{4y}}{3} + \frac{{4y}}{3}}}{5} \ge {\left[ {{{\left( {\frac{{3x}}{2}} \right)}^2}{{\left( {\frac{{4y}}{3}} \right)}^3}} \right]^{1/5}}$
$ \Rightarrow \frac{{3x + 4y}}{5} \ge {\left( {\frac{9}{4}{x^2} \times \frac{{4 \times 16}}{{9 \times 3}}{y^3}} \right)^{1/5}}$
$ \Rightarrow \frac{{10}}{5} \ge {\left( {\frac{{16}}{3}{x^2}{y^3}} \right)^{1/5}}$
$ \Rightarrow {2^5} \ge \frac{{16}}{3}{x^2}{y^3}$
$ \Rightarrow {x^2}{y^3} \le 6$
$ \Rightarrow {\left( {{x^2}{y^3}} \right)_{\max }} = 6$
$y = \frac{{{x^2} – \alpha \beta }}{{2x – \alpha – \beta }}$$x \in \mathbb{R} – \left\{ {\frac{{\alpha + \beta }}{2}} \right\}$
Assuming $\beta > \alpha $ which of the following option(s) is/are correct?
(A) $y \le \alpha $
(B) $y \ge \beta $
(C) $\alpha \le y \le \beta $
(D) $y \in \mathbb{R}$ Continue reading $y = \frac{{{x^2} – \alpha \beta }}{{2x – \alpha – \beta }}$
$x \in \mathbb{R} – \left\{ {\frac{{\alpha + \beta }}{2}} \right\}$
