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 :

<p>$\because$ ${s_1} + {s_2} = k$</p> <p>$76{x^2} + 3\pi {r^2} = k$</p> <p>$\therefore$ $152x{{dx} \over {dr}} + 6\pi r = 0$</p> <p>$\therefore$ ${{dx} \over {dr}} = {{ - 6\pi r} \over {152x}}$</p> <p>Now $V = 40{x^3} + {2 \over 3}\pi {r^3}$</p> <p>$\therefore$ ${{dv} \over {dr}} = 120{x^2}\,.\,{{dx} \over {dr}} + 2\pi {r^2} = 0$</p> <p>$$ \Rightarrow 120{x^2}\,.\,\left( {{{ - 6\pi } \over {152}}{r \over x}} \right) + 2\pi {r^2} = 0$$</p> <p>$$ \Rightarrow 120\left( {{x \over r}} \right) = 2\pi \left( {{{152} \over {6\pi }}} \right)$$</p> <p>$$ \Rightarrow \left( {{x \over r}} \right) = {{152} \over 3}{1 \over {120}} = {{19} \over {45}}$$</p>