Show that the right circular cylinder, open at the top, and of given surface area and maximum volume is such that its height is equal to the radius of the base.

Solution

let r and h be the radius and height of the right circular cylinder with the open top.

So surface area of the cylinder S is given by,

$S = π r^{2} + 2 π r h \Rightarrow h = \frac{S - π r^{2}}{2 π r} . . . . . . . . . (i)$

Let V be the volume, so

$V = π r^{2} h = π r^{2} \frac{(S - π r^{2})}{2 π r} = \frac{r (S - π r^{2})}{2} \frac{dV}{dr} = \frac{S}{2} - \frac{3 π r^{2}}{2} . . . . . . . . . . (ii) For maxima or minima \frac{dV}{dr} = 0 \Rightarrow S = 3 π r^{2} or r = \sqrt{\frac{S}{3 π}}$

Using this (i)

$h = \frac{2 π r^{2}}{2 π r} = r \frac{d^{2} V}{d r^{2}} = - 3 π r = - 3 π \sqrt{\frac{S}{3 π}} < 0 So, r = \sqrt{\frac{S}{3 π}} is a point of maxima .$

And in this case radius of base = height.

For Daily Free Study Material Join wiredfaculty

Application Of Derivatives

Show that the right circular cylinder, open at the top, and of given surface area and maximum volume is such that its height is equal to the radius of the base.

Some More Questions From Application of Derivatives Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction