CBSE 2026 · Region 5 · Set 3 · Q36 · 4 marks
A company produces cylindrical tumblers, open from the top. Since they want uniformity in the product, they fix the surface area of the tumblers produced.Based on the above information, answer the following questions : If for a tumbler, V is its volume, h the height and r the radius of the circular base, then :(i)Differentiate its volume with respect to radius of the base, where the surface area is constant.(ii)If the company wants to maximize the volume of each tumbler, then establish a relation between its height and the radius of the base. Case Study - $\displaystyle 2$
A company produces cylindrical tumblers, open from the top. Since they want uniformity in the product, they fix the surface area of the tumblers produced.
Based on the above information, answer the following questions : If for a tumbler, V is its volume, h the height and r the radius of the circular base, then :
(i)
Differentiate its volume with respect to radius of the base, where the surface area is constant.
(ii)
If the company wants to maximize the volume of each tumbler, then establish a relation between its height and the radius of the base. Case Study - $\displaystyle 2$
Marking-scheme solution
(i)
Surface area $\displaystyle S=2 \pi r h+\pi r^{2}$
$\displaystyle V=\pi r^{2} h=\dfrac{\pi r^{2}\left(S-\pi r^{2}\right)}{2 \pi r}=\dfrac{\left(S r-\pi r^{3}\right)}{2}$
$\displaystyle \dfrac{dV}{dr}=\dfrac{\left(S-3 \pi r^{2}\right)}{2}$
(ii)
$\displaystyle \dfrac{dV}{dr}=0$ gives $\displaystyle S=3 \pi r^{2}$
$\displaystyle \dfrac{d^{2} V}{dr^{2}}=-3 \pi r<0$
V is maximum when $\displaystyle S=3 \pi r^{2}$
Thus, $\displaystyle h=\dfrac{3 \pi r^{2}-\pi r^{2}}{2 \pi r}=r$
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.