Question Number 159378 by HongKing last updated on 16/Nov/21
$$\mathrm{let}\:\:\mathrm{x};\mathrm{y}>\mathrm{0}\:\:\mathrm{such}\:\mathrm{that}\:\:\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:=\:\mathrm{2} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{expression}: \\ $$$$\mathrm{P}\:=\:\mathrm{2020}\boldsymbol{\mathrm{x}}\:+\:\mathrm{2021}\boldsymbol{\mathrm{y}} \\ $$
Commented by mr W last updated on 16/Nov/21
$${P}_{{max}} \:{exists}. \\ $$$${but}\:{for}\:{x},\:{y}>\mathrm{0}\:{P}_{{min}} \:{doesn}'{t}\:{exist}. \\ $$$${only}\:{for}\:{x},\:{y}\geqslant\mathrm{0}\:{P}_{{min}} \:{exists}. \\ $$
Commented by HongKing last updated on 16/Nov/21
$$\mathrm{yes}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser},\:\mathrm{sorry}\:\mathrm{find}\:\mathrm{the}\:\mathrm{maximum} \\ $$
Answered by mr W last updated on 16/Nov/21
$${for}\:{extrem}\:{value}\:{of}\:{P}\:{the}\:{line} \\ $$$$\mathrm{2020}{x}+\mathrm{2021}{y}={P}\:{should}\:{tangent}\:{the} \\ $$$${curve}\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{2}. \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} \frac{{dy}}{{dx}}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} ×\frac{\mathrm{2020}}{\mathrm{2021}}=\mathrm{0} \\ $$$$\Rightarrow{y}=\sqrt{\frac{\mathrm{2021}}{\mathrm{2020}}}{x} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{2021}}{\mathrm{2020}}\sqrt{\frac{\mathrm{2021}}{\mathrm{2020}}}{x}^{\mathrm{3}} =\mathrm{2} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{2021}}{\mathrm{2020}}\sqrt{\frac{\mathrm{2021}}{\mathrm{2020}}}}} \\ $$$${y}=\sqrt{\frac{\mathrm{2021}}{\mathrm{2020}}}\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{2021}}{\mathrm{2020}}\sqrt{\frac{\mathrm{2021}}{\mathrm{2020}}}}} \\ $$$${P}_{{max}} =\left(\mathrm{2020}+\mathrm{2021}\sqrt{\frac{\mathrm{2021}}{\mathrm{2020}}}\right)\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{2021}}{\mathrm{2020}}\sqrt{\frac{\mathrm{2021}}{\mathrm{2020}}}}} \\ $$$$=\mathrm{2020}\sqrt[{\mathrm{3}}]{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{2021}}{\mathrm{2020}}\sqrt{\frac{\mathrm{2021}}{\mathrm{2020}}}\right)^{\mathrm{2}} } \\ $$$$\approx\mathrm{4041}.\mathrm{000062} \\ $$
Commented by HongKing last updated on 16/Nov/21
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser}\:\mathrm{perfect} \\ $$
Commented by mr W last updated on 16/Nov/21
