Question Number 84220 by Rio Michael last updated on 10/Mar/20
$$\mathrm{find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{2}}{\mathrm{3cosh2}{x}\:+\mathrm{2}} \\ $$
Commented by mr W last updated on 10/Mar/20
$$\mathrm{cosh}\:\mathrm{2}{x}\:\geqslant\mathrm{1} \\ $$$$\mathrm{3}\:\mathrm{cosh}\:\mathrm{2}{x}+\mathrm{2}\:\geqslant\mathrm{3}×\mathrm{1}+\mathrm{2}=\mathrm{5} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}\:\mathrm{cosh}\:\mathrm{2}{x}+\mathrm{2}}\leqslant\frac{\mathrm{2}}{\mathrm{5}}\:={maximum} \\ $$
Answered by TANMAY PANACEA last updated on 10/Mar/20
$$\frac{\mathrm{2}}{\mathrm{3}\left(\frac{{e}^{\mathrm{2}{x}} +{e}^{−\mathrm{2}{x}} }{\mathrm{2}}\right)+\mathrm{2}}={y}\: \\ $$$${y}=\frac{\mathrm{4}}{\mathrm{3}{e}^{\mathrm{2}{x}} +\mathrm{3}{e}^{−\mathrm{2}{x}} +\mathrm{4}}=\frac{\mathrm{4}}{\mathrm{3}{a}+\frac{\mathrm{3}}{{a}}+\mathrm{4}}=\frac{\mathrm{4}}{\mathrm{3}\left\{\left(\sqrt{{a}}\:−\frac{\mathrm{1}}{\:\sqrt{{a}}}\right)^{\mathrm{2}} +\mathrm{2}\right\}+\mathrm{4}} \\ $$$${y}_{{max}} =\frac{\mathrm{4}}{\mathrm{3}\left(\mathrm{0}+\mathrm{2}\right)+\mathrm{4}}=\frac{\mathrm{4}}{\mathrm{10}}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$
Commented by Rio Michael last updated on 10/Mar/20
$${thanks} \\ $$