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

$If \frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} [R_{1}, R_{2} > 0] and$ $R_{1} + R_{2} = C (Constant) then prove that$ $R will be maximum when R_{1} = R_{2} .$
	Answered by mr W last updated on 07/Jun/24

	$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{R_{1} + R_{2}}{R_{1} R_{2}} = \frac{C}{R_{1} R_{2}}$ $R = \frac{R_{1} R_{2}}{C} = \frac{{(\sqrt{R_{1} R_{2}})}^{2}}{C} ⩽ \frac{1}{C} {(\frac{R_{1} + R_{2}}{2})}^{2} = \frac{C}{4}$ $R_{m a x} = \frac{C}{4}$ $E q u a l i t y w h e n R_{1} = R_{2}$
	Answered by zamin2001 last updated on 07/Jun/24

	$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} \Rightarrow R = \frac{R_{1} \times R_{2}}{R_{1} + R_{2}} & R_{1} + R_{2} = C & R_{2} = C - R_{1}$ $\Rightarrow R = \frac{R_{1} \times (C - R_{1})}{C} = R_{1} - \frac{1}{C} R_{1}^{2} \Rightarrow R^{^{'}} = 1 - \frac{2}{C} R_{1}$ $\overset{R^{'} = 0}{\Rightarrow} 1 - \frac{2}{C} R_{1} = 0 \Rightarrow R_{1} = \frac{C}{2} & R_{2} = C - R_{1} = C - \frac{C}{2} = \frac{C}{2}$ $\Rightarrow R_{1} = R_{2}$
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