Question Number 198434 by cortano12 last updated on 20/Oct/23
Commented by mr W last updated on 20/Oct/23
$${due}\:{to}\:{symmetry} \\ $$$${max}=\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$
Answered by mr W last updated on 20/Oct/23
Commented by mr W last updated on 20/Oct/23
$${general}\:{case} \\ $$$${c}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${y}={c}−{x} \\ $$$$\mathrm{cos}\:\theta=\frac{{b}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$${z}=\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}{ax}\:\mathrm{cos}\:\theta} \\ $$$$\:=\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} −\frac{\mathrm{2}{abx}}{\:{c}}} \\ $$$${P}={xyz}={x}\left({c}−{x}\right)\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} −\frac{\mathrm{2}{abx}}{{c}}} \\ $$$${let}\:{a}=\alpha{c},\:{b}=\beta{c},\:{x}=\xi{c} \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\mathrm{1} \\ $$$$\Phi=\frac{{P}}{{c}^{\mathrm{3}} }=\xi\left(\mathrm{1}−\xi\right)\sqrt{\xi^{\mathrm{2}} +\alpha^{\mathrm{2}} −\mathrm{2}\alpha\beta\xi} \\ $$$$\Phi=\frac{{P}}{{c}^{\mathrm{3}} }=\xi\left(\mathrm{1}−\xi\right)\sqrt{\left(\xi−\alpha\beta\right)^{\mathrm{2}} +\alpha^{\mathrm{4}} } \\ $$$$\frac{{d}\Phi}{{d}\xi}=…=\mathrm{0} \\ $$$$… \\ $$