If-1-R-1-R-1-1-R-2-R-1-R-2-gt-0-and-R-1-R-2-C-Constant-then-prove-that-R-will-be-maximum-when-R-1-R-2-

Question Number 208215 by MATHEMATICSAM last updated on 07/Jun/24

$$\mathrm{If}\:\frac{\mathrm{1}}{\mathrm{R}}\:=\:\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{1}} }\:+\:\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{2}} }\:\left[\mathrm{R}_{\mathrm{1}} ,\:\mathrm{R}_{\mathrm{2}} \:>\:\mathrm{0}\right]\:\mathrm{and}\: \\ $$$$\mathrm{R}_{\mathrm{1}} \:+\:\mathrm{R}_{\mathrm{2}} \:=\:\mathrm{C}\:\left(\mathrm{Constant}\right)\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{R}\:\mathrm{will}\:\mathrm{be}\:\mathrm{maximum}\:\mathrm{when}\:\mathrm{R}_{\mathrm{1}} \:=\:\mathrm{R}_{\mathrm{2}} . \\ $$

Answered by mr W last updated on 07/Jun/24

$$\frac{\mathrm{1}}{{R}}=\frac{\mathrm{1}}{{R}_{\mathrm{1}} }+\frac{\mathrm{1}}{{R}_{\mathrm{2}} }=\frac{{R}_{\mathrm{1}} +{R}_{\mathrm{2}} }{{R}_{\mathrm{1}} {R}_{\mathrm{2}} }=\frac{{C}}{{R}_{\mathrm{1}} {R}_{\mathrm{2}} } \\ $$$${R}=\frac{{R}_{\mathrm{1}} {R}_{\mathrm{2}} }{{C}}=\frac{\left(\sqrt{{R}_{\mathrm{1}} {R}_{\mathrm{2}} }\right)^{\mathrm{2}} }{{C}}\leqslant\frac{\mathrm{1}}{{C}}\left(\frac{{R}_{\mathrm{1}} +{R}_{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} =\frac{{C}}{\mathrm{4}} \\ $$$${R}_{{max}} =\frac{{C}}{\mathrm{4}} \\ $$$${Equality}\:{when}\:{R}_{\mathrm{1}} ={R}_{\mathrm{2}} \\ $$

Answered by zamin2001 last updated on 07/Jun/24

$$\frac{\mathrm{1}}{{R}}=\frac{\mathrm{1}}{{R}_{\mathrm{1}} }+\frac{\mathrm{1}}{{R}_{\mathrm{2}} }\Rightarrow{R}=\frac{{R}_{\mathrm{1}} ×{R}_{\mathrm{2}} }{{R}_{\mathrm{1}} +{R}_{\mathrm{2}} }\:\:\&\:{R}_{\mathrm{1}} +{R}_{\mathrm{2}} ={C}\:\&{R}_{\mathrm{2}} ={C}−{R}_{\mathrm{1}} \\ $$$$\Rightarrow\:{R}=\frac{{R}_{\mathrm{1}} ×\left({C}−{R}_{\mathrm{1}} \right)}{{C}}={R}_{\mathrm{1}} −\frac{\mathrm{1}}{{C}}{R}_{\mathrm{1}} ^{\mathrm{2}} \Rightarrow{R}^{'} =\mathrm{1}−\frac{\mathrm{2}}{{C}}{R}_{\mathrm{1}} \\ $$$$\overset{{R}'=\mathrm{0}} {\Rightarrow}\:\mathrm{1}−\frac{\mathrm{2}}{{C}}{R}_{\mathrm{1}} =\mathrm{0}\Rightarrow\:{R}_{\mathrm{1}} =\frac{{C}}{\mathrm{2}}\:\&\:{R}_{\mathrm{2}} ={C}−{R}_{\mathrm{1}} ={C}−\frac{{C}}{\mathrm{2}}=\frac{{C}}{\mathrm{2}} \\ $$$$\Rightarrow\:{R}_{\mathrm{1}} ={R}_{\mathrm{2}} \\ $$

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