Question Number 53769 by ajfour last updated on 25/Jan/19
$${Find}\:{maximum}\:{of} \\ $$$$\:\:\:\frac{{xyz}}{\left({x}+{a}\right)\left({x}+{y}\right)\left({y}+{z}\right)\left({z}+{b}\right)}\:\:. \\ $$
Commented by mr W last updated on 25/Jan/19
$$\frac{\mathrm{1}}{\mathrm{16}\sqrt{{ab}}}\:? \\ $$
Commented by ajfour last updated on 25/Jan/19
$$\left({a}^{\mathrm{1}/\mathrm{4}} +{b}^{\mathrm{1}/\mathrm{4}} \right)^{−\mathrm{4}} \:\:\:{Sir}. \\ $$
Commented by ajfour last updated on 25/Jan/19
$${long}\:{way},\:{Sir}.\:{i}\:{got}\:{it}\:{just}\:{now} \\ $$$$\frac{\partial{f}}{\partial{x}}=\mathrm{0}\:\:\:\Rightarrow\:\:{x}^{\mathrm{2}} ={ay} \\ $$$$\frac{\partial{f}}{\partial{y}}=\:\mathrm{0}\:\:\:\Rightarrow\:\:{y}^{\mathrm{2}} ={xz} \\ $$$$\frac{\partial{f}}{\partial{z}}=\mathrm{0}\:\:\:\Rightarrow\:\:{z}^{\mathrm{2}} ={by} \\ $$$${hence}\:\:\:{xz}={ab} \\ $$$${Then}\:\:{x}^{\mathrm{2}} ={a}\sqrt{{ab}}\:\:,\:{y}^{\mathrm{2}} =\sqrt{{ab}}\:,\:{z}^{\mathrm{2}} ={b}\sqrt{{ab}} \\ $$$${after}\:{much}\:{length}\:{i}\:{could}\:{get} \\ $$$$\:\:{f}\left({x},{y},{z}\right)\:{is}\:{then}\:=\:\left({a}^{\mathrm{1}/\mathrm{4}} +{b}^{\mathrm{1}/\mathrm{4}} \right)^{−\mathrm{4}} \:. \\ $$
Commented by mr W last updated on 26/Jan/19
$${you}\:{are}\:{right}\:{sir}! \\ $$$${i}\:{considered}\:{only}\:{the}\:{case}\:{a}={b}. \\ $$