Question

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 :

• A. $2:5$
• B. $19:45$
• C. $3:8$
• D. $19:15$

Answer 1

Answer: (B)

Surface area $=76x^2+3\pi r^2=$ constant $(K)$
$V=40x^3+\frac{2}{3}\pi r^3$
$\left[76x^2+3\pi r^2=K\right]$
$r^2=\frac{K-76x^2}{3\pi }$
$r={\left(\frac{K-76x^2}{3\pi }\right)}^{\frac{1}{2}}$
$V=40x^3+\frac{2}{3}\pi {\left(\frac{K-76x^2}{3\pi }\right)}^{\frac{3}{2}}$
$\frac{dV}{dx}=120x^2+\frac{2}{3}\pi \cdot \frac{3}{2}{\left(\frac{K-76x^2}{3\pi }\right)}^{\frac{1}{2}}\cdot \left(\frac{-76(2x)}{3\pi }\right)$
Put
$\frac{dV}{dx}=0\Rightarrow 120x^2+\frac{2}{3}\pi \cdot \frac{3}{2}{\left(\frac{K-76x^2}{3\pi }\right)}^{\frac{1}{2}}\cdot \left(\frac{-76(2x)}{3\pi }\right)=0$
$\Rightarrow 120x^2=\frac{152x}{3}{\left(\frac{k-76x^2}{3\pi }\right)}^{\frac{1}{2}}$
$\Rightarrow \frac{45}{19}{x}^2=x{\left(\frac{k-76x^2}{3\pi }\right)}^{\frac{1}{2}};x\ne 0$
$\Rightarrow \frac{45}{19}x={\left(\frac{k-76x^2}{3\pi }\right)}^{\frac{1}{2}}$
$\Rightarrow {\left(\frac{45}{19}\right)}^2{x}^2=\frac{k-76x^2}{3\pi }$
$\Rightarrow {\left(\frac{45}{19}\right)}^2{x}^2=r^2$
$\Rightarrow \frac{x^2}{r^2}={\left(\frac{19}{45}\right)}^2$
$\Rightarrow \frac{x}{r}=\frac{19}{45}$