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Cosider a cuboid of sides \mathrm{2 x, 4 x\text{ and }5 x} and a closed hemisphere of radius \mathrm{r}. If the sum of their surface areas is a constant \mathrm{k}, then the ratio \mathrm{x:r}, for which the sum of their volumes is maximum, is:

Option: 1

2:5


Option: 2

19:45


Option: 3

3:8


Option: 4

19:15


Answers (1)

best_answer

\begin{aligned} &S=2\left(8 x^{2}+20 x^{2}+10 x^{2}\right)+\left(2 \pi r^{2}+\pi r^{2}\right) \\ &K=76 x^{2}+3 \pi r^{2} \\ &\text { Volume }(V)=40 x^{3}+\frac{2}{3} \pi r^{3} \\ \end{aligned}

\begin{aligned} &V(r)=40\left(\frac{k-3 \pi r^{2}}{76}\right)^{3 / 2}+\frac{2}{3} \pi r^{3} \\ &V^{\prime}(r)=40\left(\frac{3}{2}\right)\left(\frac{k-3 \pi r^{2}}{76}\right)^{1 / 2}\left(-\frac{3 \pi}{30} r\right)+2 \pi r^{2} \end{aligned}

v^{\prime}(r)=0\\ 60\left(\frac{3 \pi r}{38}\right)\left(\frac{k-3 \pi r^{2}}{76}\right)^{1 / 2}=2 \pi r^{2}\\\begin{aligned} \frac{90}{19} \quad x &=2 n \\ \frac{x}{x} &=\frac{19}{45} \end{aligned}

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Divya Prakash Singh

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