JEE Main 2022 (Online) 26th June Evening ShiftQuestion 1
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 :
Differentiation and its Application
Solution
Concept. Treat $r$ as a function of $x$ through the constraint; differentiate the constraint and the volume in $x$. Formulas used. (1) Cuboid: surface $2(8+10+20)x^2=76x^2$, volume $40x^3$(2) Closed hemisphere: surface $2\pi r^2+\pi r^2=3\pi r^2$, volume $\frac23\pi r^3$Step 1 - Constraint. By (1), (2): $76x^2+3\pi r^2=k$. Differentiating in $x$: $$152x+6\pi r\,\frac{dr}{dx}=0\ \Longrightarrow\ \frac{dr}{dx}=-\frac{76x}{3\pi r}.$$ Step 2 - Volume. $V=40x^3+\frac23\pi r^3$, so $$\frac{dV}{dx}=120x^2+2\pi r^2\frac{dr}{dx}=120x^2-\frac{152}{3}xr.$$ Step 3 - Critical point. $\frac{dV}{dx}=0$ gives $120x=\frac{152}{3}r$, i.e. $r=\frac{360x}{152}=\frac{45x}{19}$. Step 4 - Ratio. $x:r=19:45$. Answer: (B)