A straight line AB of length 8cm is divided into two parts AP and PB by point P. Find the position of P if AP^2 + BP^2 is minimum

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$$\begin{gathered} \mathrm{Let} \mathrm{AP}=\mathrm{x} \mathrm{and} \mathrm{BP}=\mathrm{y} \\ \Rightarrow \mathrm{x}+\mathrm{y}=8 \\ \Rightarrow \mathrm{y}=8-\mathrm{x} \\ \mathrm{Now} \\ \mathrm{Let} \mathrm{S}={\left(\mathrm{AP}\right)}^2+{\left(\mathrm{BP}\right)}^2 \\ \Rightarrow \mathrm{S}={\mathrm{x}}^2+{\mathrm{y}}^2 \\ \Rightarrow \mathrm{S}={\mathrm{x}}^2+{\left(8-\mathrm{x}\right)}^2 \\ \mathrm{Differentiating} \mathrm{both} \mathrm{sides} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \mathrm{we} \mathrm{get}, \\ \frac{\mathrm{dS}}{\mathrm{dx}}=2\mathrm{x}+2\left(8-\mathrm{x}\right)\left(-1\right) \\ \Rightarrow \frac{\mathrm{dS}}{\mathrm{dx}}=2\mathrm{x}-2\left(8-\mathrm{x}\right)=2\mathrm{x}-16+2\mathrm{x} \\ \Rightarrow \frac{\mathrm{dS}}{\mathrm{dx}}=4\mathrm{x}-16 \\ \mathrm{For} \mathrm{maixima} \mathrm{or} \mathrm{minima} \frac{\mathrm{dS}}{\mathrm{dx}}=0 \\ \Rightarrow 4\mathrm{x}-16=0 \\ \Rightarrow \mathrm{x}=4 \\ \Rightarrow \frac{{\mathrm{d}}^2\mathrm{S}}{{\mathrm{dx}}^2}=4 \mathrm{which} \mathrm{is} \mathrm{positive}. \mathrm{Hence} \mathrm{S} \mathrm{will} \mathrm{be} \mathrm{minimum} \mathrm{at} \mathrm{x}=4. \\ \Rightarrow \mathrm{x}=4 \\ \Rightarrow \mathrm{y}=8-\mathrm{x}=8-4 \\ \Rightarrow \mathrm{y}=4 \\ \mathbf{Hence}\mathbf{ }\mathbf{P}\mathbf{ }\mathbf{will}\mathbf{ }\mathbf{be}\mathbf{ }\mathbf{at}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{mid}\mathbf{ }\mathbf{point}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{AB}\mathbf{.} \end{gathered}$$

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