Explanation:
\({I_{\max }} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
According to question
\(\frac{{{I_1}}}{{{I_2}}} = \frac{2}{8} = \frac{1}{4}\;\;\;{\mkern 1mu} {\kern 1pt} \therefore \;\;\;{\mkern 1mu} {\kern 1pt} {I_2} = 4{I_1}{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\)
From eqs. (1) and (2)
\({I_{\max }} = {I_1} + 4{I_1} + 2\sqrt {4I_1^2} = 5{I_1} + 4{I_1}\)
\({I_{\max }} = 9{I_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 3 \right)\)
\({I_{\min }} = {I_1} + {I_2} - 2\sqrt {4{I_1}{I_2}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 4 \right)\)
From eqs. (2) and (4)
\({I_{\min {\rm{ }}}} = {I_1} + 4{I_1} - 2\sqrt {4I_1^2} \;\;\;{\mkern 1mu} {\kern 1pt} {I_{\min {\rm{ }}}} = {I_1}\)