QJEEM22Ma2f51: JEE Main 2022 Mathematics - Class 12 JEEM22Ma2f51 Solution: Consider a cuboid of sides 2x, 4x and 5x and a closed…

$\because$ ${s_1} + {s_2} = k$

$76{x^2} + 3\pi {r^2} = k$

$\therefore$ $152x{{dx} \over {dr}} + 6\pi r = 0$

$\therefore$ ${{dx} \over {dr}} = {{ - 6\pi r} \over {152x}}$

Now $V = 40{x^3} + {2 \over 3}\pi {r^3}$

$\therefore$ ${{dv} \over {dr}} = 120{x^2}\,.\,{{dx} \over {dr}} + 2\pi {r^2} = 0$

$\Rightarrow 120{x^2}\,.\,\left( {{{ - 6\pi } \over {152}}{r \over x}} \right) + 2\pi {r^2} = 0$

$\Rightarrow 120\left( {{x \over r}} \right) = 2\pi \left( {{{152} \over {6\pi }}} \right)$

$\Rightarrow \left( {{x \over r}} \right) = {{152} \over 3}{1 \over {120}} = {{19} \over {45}}$