If a cylindrical vessel of given volume V wit | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given that the volume of the cylindrical vessel is $V$ and it has no lid on the top.Let $r$ be the radius and $h$ be the height of the cylinder.Th

Vedclass · Accepted Answer

$r=\sqrt[3]{\frac{V}{\pi}}, h=\sqrt[3]{\frac{V}{\pi}}$

Vedclass · Answer

(C) Given that the volume of the cylindrical vessel is $V$ and it has no lid on the top.Let $r$ be the radius and $h$ be the height of the cylinder.The volume of the cylinder is given by $V = \pi r^2 h$,which implies $h = \frac{V}{\pi r^2}$.The surface area $S$ of the metal sheet used (including the base but no lid) is given by:$S = 2\pi rh + \pi r^2$Substituting $h = \frac{V}{\pi r^2}$ into the expression for $S$:$S(r) = 2\pi r \left(\frac{V}{\pi r^2}\right) + \pi r^2 = \frac{2V}{r} + \pi r^2$To find the minimum surface area,we differentiate $S(r)$ with respect to $r$ and set it to zero:$S'(r) = -\frac{2V}{r^2} + 2\pi r = 0$$2\pi r = \frac{2V}{r^2} \Rightarrow r^3 = \frac{V}{\pi} \Rightarrow r = \sqrt[3]{\frac{V}{\pi}}$Now,substitute $r$ back into the expression for $h$:$h = \frac{V}{\pi r^2} = \frac{V}{\pi (V/\pi)^{2/3}} = \frac{V}{\pi} \cdot \left(\frac{\pi}{V}\right)^{2/3} = \left(\frac{V}{\pi}\right)^{1/3} = \sqrt[3]{\frac{V}{\pi}}$Thus,the radius and height for minimum surface area are $r = \sqrt[3]{\frac{V}{\pi}}$ and $h = \sqrt[3]{\frac{V}{\pi}}$.

If a cylindrical vessel of given volume $V$ with no lid on the top is to be made from a sheet of metal,then the radius $(r)$ and height $(h)$ of the vessel so that the metal sheet used is minimum,is

Similar Questions

If ${a^2}{x^4} + {b^2}{y^4} = {c^6}$,then the maximum value of $xy$ is

The function $f(x) = x^5 - 5x^4 + 5x^3 - 10$ has a local maximum when $x$ is equal to:

The minimum value of $\alpha$ for which the equation $\frac{4}{\sin x}+\frac{1}{1-\sin x}=\alpha$ has at least one solution in $\left(0, \frac{\pi}{2}\right)$ is ..........

The number $10$ is divided into two parts such that the sum of double of the first part and the square of the second part is minimum. The two parts are respectively:

Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $R$ is $\frac{2 R}{\sqrt{3}} .$ Also find the maximum volume.

Vedclass Test Series

Exam Paper Generator

Online Exam Module