JEE MAIN - Mathematics (2022 - 26th June Evening Shift - No. 5)
Consider a cuboid of sides 2x, 4x and 5x and a closed hemisphere of radius r. If the sum of their surface areas is a constant k, then the ratio x : r, for which the sum of their volumes is maximum, is :
2 : 5
19 : 45
3 : 8
19 : 15
Explanation
$$\because$$ $${s_1} + {s_2} = k$$
$$76{x^2} + 3\pi {r^2} = k$$
$$\therefore$$ $$152x{{dx} \over {dr}} + 6\pi r = 0$$
$$\therefore$$ $${{dx} \over {dr}} = {{ - 6\pi r} \over {152x}}$$
Now $$V = 40{x^3} + {2 \over 3}\pi {r^3}$$
$$\therefore$$ $${{dv} \over {dr}} = 120{x^2}\,.\,{{dx} \over {dr}} + 2\pi {r^2} = 0$$
$$ \Rightarrow 120{x^2}\,.\,\left( {{{ - 6\pi } \over {152}}{r \over x}} \right) + 2\pi {r^2} = 0$$
$$ \Rightarrow 120\left( {{x \over r}} \right) = 2\pi \left( {{{152} \over {6\pi }}} \right)$$
$$ \Rightarrow \left( {{x \over r}} \right) = {{152} \over 3}{1 \over {120}} = {{19} \over {45}}$$
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