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Question Number 159378 by HongKing last updated on 16/Nov/21

$$\mathrm{let}\mathrm{x};\mathrm{y}>0\mathrm{such}\mathrm{that}{\mathrm{x}}^3+{\mathrm{y}}^3=2$$ $$\mathrm{find}\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}\mathrm{the}$$ $$\mathrm{following}\mathrm{expression}:$$ $$\mathrm{P}=2020\mathbf{x}+2021\mathbf{y}$$

Commented bymr W last updated on 16/Nov/21

$$P_{max}exists.$$ $$butforx,y>0P_{min}{doesn}^{\prime }texist.$$ $$onlyforx,y⩾0P_{min}exists.$$

Commented byHongKing 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

$$forextremvalueofPtheline$$ $$2020x+2021y=Pshouldtangentthe$$ $$curve{x}^3+y^3=2.$$ $$2x^2+3y^2\frac{dy}{dx}=0$$ $$x^2-y^2\times \frac{2020}{2021}=0$$ $$\Rightarrow y=\sqrt{\frac{2021}{2020}}x$$ $$x^3+\frac{2021}{2020}\sqrt{\frac{2021}{2020}}{x}^3=2$$ $$x=\sqrt[3]{\frac{2}{1+\frac{2021}{2020}\sqrt{\frac{2021}{2020}}}}$$ $$y=\sqrt{\frac{2021}{2020}}\sqrt[3]{\frac{2}{1+\frac{2021}{2020}\sqrt{\frac{2021}{2020}}}}$$ $$P_{max}=(2020+2021\sqrt{\frac{2021}{2020}})\sqrt[3]{\frac{2}{1+\frac{2021}{2020}\sqrt{\frac{2021}{2020}}}}$$ $$=2020\sqrt[3]{2{(1+\frac{2021}{2020}\sqrt{\frac{2021}{2020}})}^2}$$ $$\approx 4041.000062$$

Commented byHongKing last updated on 16/Nov/21

$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{my}\mathrm{dear}\mathrm{Ser}\mathrm{perfect}$$

Commented bymr W last updated on 16/Nov/21

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