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let-x-y-gt-0-such-that-x-3-y-3-2-find-the-minimum-value-of-the-following-expression-P-2020x-2021y-

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Question Number 159378 by HongKing last updated on 16/Nov/21
let x;y>0 such that x^3 + y^3 = 2 find the minimum value of the following expression: P = 2020x + 2021y
$$\mathrm{let}\:\:\mathrm{x};\mathrm{y}>\mathrm{0}\:\:\mathrm{such}\:\mathrm{that}\:\:\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:=\:\mathrm{2} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{expression}: \\ $$$$\mathrm{P}\:=\:\mathrm{2020}\boldsymbol{\mathrm{x}}\:+\:\mathrm{2021}\boldsymbol{\mathrm{y}} \\ $$
Commented by mr W last updated on 16/Nov/21
P_(max) exists. but for x, y>0 P_(min) doesn′t exist. only for x, y≥0 P_(min) exists.
$${P}_{{max}} \:{exists}. \\ $$$${but}\:{for}\:{x},\:{y}>\mathrm{0}\:{P}_{{min}} \:{doesn}'{t}\:{exist}. \\ $$$${only}\:{for}\:{x},\:{y}\geqslant\mathrm{0}\:{P}_{{min}} \:{exists}. \\ $$
Commented by HongKing last updated on 16/Nov/21
yes my dear Ser, sorry find the maximum
$$\mathrm{yes}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser},\:\mathrm{sorry}\:\mathrm{find}\:\mathrm{the}\:\mathrm{maximum} \\ $$
Answered by mr W last updated on 16/Nov/21
for extrem value of P the line 2020x+2021y=P should tangent the curve x^3 +y^3 =2. 2x^2 +3y^2 (dy/dx)=0 x^2 −y^2 ×((2020)/(2021))=0 ⇒y=(√((2021)/(2020)))x x^3 +((2021)/(2020))(√((2021)/(2020)))x^3 =2 x=((2/(1+((2021)/(2020))(√((2021)/(2020))))))^(1/3) y=(√((2021)/(2020)))((2/(1+((2021)/(2020))(√((2021)/(2020))))))^(1/3) P_(max) =(2020+2021(√((2021)/(2020))))((2/(1+((2021)/(2020))(√((2021)/(2020))))))^(1/3) =2020((2(1+((2021)/(2020))(√((2021)/(2020))))^2 ))^(1/3) ≈4041.000062
$${for}\:{extrem}\:{value}\:{of}\:{P}\:{the}\:{line} \\ $$$$\mathrm{2020}{x}+\mathrm{2021}{y}={P}\:{should}\:{tangent}\:{the} \\ $$$${curve}\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{2}. \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} \frac{{dy}}{{dx}}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} ×\frac{\mathrm{2020}}{\mathrm{2021}}=\mathrm{0} \\ $$$$\Rightarrow{y}=\sqrt{\frac{\mathrm{2021}}{\mathrm{2020}}}{x} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{2021}}{\mathrm{2020}}\sqrt{\frac{\mathrm{2021}}{\mathrm{2020}}}{x}^{\mathrm{3}} =\mathrm{2} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{2021}}{\mathrm{2020}}\sqrt{\frac{\mathrm{2021}}{\mathrm{2020}}}}} \\ $$$${y}=\sqrt{\frac{\mathrm{2021}}{\mathrm{2020}}}\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{2021}}{\mathrm{2020}}\sqrt{\frac{\mathrm{2021}}{\mathrm{2020}}}}} \\ $$$${P}_{{max}} =\left(\mathrm{2020}+\mathrm{2021}\sqrt{\frac{\mathrm{2021}}{\mathrm{2020}}}\right)\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{2021}}{\mathrm{2020}}\sqrt{\frac{\mathrm{2021}}{\mathrm{2020}}}}} \\ $$$$=\mathrm{2020}\sqrt[{\mathrm{3}}]{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{2021}}{\mathrm{2020}}\sqrt{\frac{\mathrm{2021}}{\mathrm{2020}}}\right)^{\mathrm{2}} } \\ $$$$\approx\mathrm{4041}.\mathrm{000062} \\ $$
Commented by HongKing last updated on 16/Nov/21
thank you so much my dear Ser perfect
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser}\:\mathrm{perfect} \\ $$
Commented by mr W last updated on 16/Nov/21

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