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Q.
Consider a cuboid of sides $2 x , 4 x$ and $5 x$ 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 :
Surface area $=76 x ^{2}+3 \pi r ^{2}=$ constant $( K )$
$V =40 x ^{3}+\frac{2}{3} \pi r ^{3}$
${\left[76 x ^{2}+3 \pi r ^{2}= K \right]}$
$r ^{2}=\frac{ K -76 x ^{2}}{3 \pi}$
$r =\left(\frac{ K -76 x ^{2}}{3 \pi}\right)^{\frac{1}{2}}$
$V =40 x ^{3}+\frac{2}{3} \pi\left(\frac{ K -76 x ^{2}}{3 \pi}\right)^{\frac{3}{2}}$
$\frac{ dV }{ dx }=120 x ^{2}+\frac{2}{3} \pi \cdot \frac{3}{2}\left(\frac{ K -76 x ^{2}}{3 \pi}\right)^{\frac{1}{2}} \cdot\left(\frac{-76(2 x )}{3 \pi}\right)$
Put
$\frac{ dV }{ dx } =0 \Rightarrow 120 x ^{2}+\frac{2}{3} \pi \cdot \frac{3}{2}\left(\frac{ K -76 x ^{2}}{3 \pi}\right)^{\frac{1}{2}} \cdot\left(\frac{-76(2 x )}{3 \pi}\right)=0$
$\Rightarrow 120 x ^{2}=\frac{152 x }{3}\left(\frac{ k -76 x ^{2}}{3 \pi}\right)^{\frac{1}{2}}$
$\Rightarrow \frac{45}{19} x ^{2}= x \left(\frac{ k -76 x ^{2}}{3 \pi}\right)^{\frac{1}{2}} ; x \neq 0$
$\Rightarrow \frac{45}{19} x =\left(\frac{ k -76 x ^{2}}{3 \pi}\right)^{\frac{1}{2}}$
$\Rightarrow\left(\frac{45}{19}\right)^{2} x ^{2}=\frac{ k -76 x ^{2}}{3 \pi}$
$\Rightarrow\left(\frac{45}{19}\right)^{2} x ^{2}= r ^{2}$
$\Rightarrow \frac{ x ^{2}}{ r ^{2}}=\left(\frac{19}{45}\right)^{2}$
$\Rightarrow \frac{ x }{ r }=\frac{19}{45}$